[proofplan] For $K_0$, we use the standard Morita equivalence defined by the idempotent matrix unit $e_{11}\in M_n(R)$ and check that it preserves finite direct sums and finite projectivity. This identifies the commutative monoids of isomorphism classes of finitely generated projective modules, so their group completions are isomorphic. For $K_1$, we use the convention $K_1(A)=GL(A)/E(A)$ for a unital ring $A$, expand matrices over $M_n(R)$ into block matrices over $R$, and compare the corresponding stable elementary subgroups. Naturality follows because both the Morita functors and the block expansion maps commute with applying a unital ring homomorphism entrywise. [/proofplan]

[step:Identify finitely generated projectives under the matrix-ring Morita equivalence]Put $A:=M_n(R)$. For $1\le i,j\le n$, let $e_{ij}\in A$ denote the matrix unit with $1_R$ in the $(i,j)$-entry and $0_R$ elsewhere, and put $e:=e_{11}$. The hypothesis $n\ge 1$ ensures that these matrix units, and in particular $e_{11}$, are defined. The corner ring $eAe$ is identified with $R$ by the ring isomorphism \begin{align*} R \to eAe, \qquad r \mapsto \text{the matrix whose }(1,1)\text{-entry is }r\text{ and whose other entries are }0_R. \end{align*} Let $\operatorname{Mod}(A)$ denote the category of left $A$-modules and $A$-linear maps, and let $\operatorname{Mod}(R)$ denote the category of left $R$-modules and $R$-linear maps. Define a functor $F:\operatorname{Mod}(A)\to\operatorname{Mod}(R)$ by $F(M)=eM$ on left $A$-modules and by restriction on homomorphisms. Define a functor $G:\operatorname{Mod}(R)\to\operatorname{Mod}(A)$ by $G(N)=Ae\otimes_R N$ on left $R$-modules. Here $Ae$ is regarded as an $(A,R)$-bimodule through the corner identification $eAe\cong R$. The standard Morita identities give natural isomorphisms \begin{align*} F(G(N)) \cong N \end{align*} for every left $R$-module $N$, and \begin{align*} G(F(M)) \cong M \end{align*} for every left $A$-module $M$. Explicitly, the first isomorphism sends \begin{align*} e(ae\otimes x) \mapsto eae\cdot x, \end{align*} where $a\in A$ and $x\in N$. Its inverse sends $x\in N$ to $e(e\otimes x)\in e(Ae\otimes_R N)$. The two composites are the identity because $eAe\cong R$ and $e$ corresponds to $1_R$. The second isomorphism is induced by the action map $\mu_M:Ae\otimes_R eM\to M$ sending $ae\otimes y$ to $ay$. Since $1_A=\sum_{j=1}^{n} e_{j1}e_{1j}$, this action map is surjective. Define $\nu_M:M\to Ae\otimes_R eM$ by $\nu_M(m)=\sum_{j=1}^{n} e_{j1}\otimes e_{1j}m$. The element $e_{1j}m$ lies in $eM$ because $e e_{1j}=e_{1j}$. A direct computation using $e_{ab}e_{cd}=\delta_{bc}e_{ad}$ gives \begin{align*} \mu_M\circ\nu_M=\operatorname{id}_M. \end{align*} It also gives \begin{align*} \nu_M\circ\mu_M=\operatorname{id}_{Ae\otimes_R eM}. \end{align*} Hence $F$ and $G$ are inverse equivalences of module categories. Both $F$ and $G$ preserve finite direct sums. The functor $F$ sends the finite free module $A^q$ to $(eA)^q$. Under $eAe\cong R$, the left $R$-module $eA$ is isomorphic to $R^n$ by reading off the first row, so $(eA)^q$ is finite free, hence finitely generated projective. The functor $G$ sends the finite free module $R^q$ to $(Ae)^q$, and $Ae$ is a direct summand of the finite free left $A$-module $A$ because $A=Ae\oplus A(1_A-e)$. Therefore both functors carry direct summands of finite free modules to finitely generated projective modules. Consequently they restrict to inverse equivalences between finitely generated projective left $A$-modules and finitely generated projective left $R$-modules.[/step]

[guided]The point of Morita theory here is that a module over $M_n(R)$ contains all its information in the first coordinate selected by the idempotent $e=e_{11}$. We make this precise. Let $A:=M_n(R)$. For $1\le i,j\le n$, let $e_{ij}\in A$ denote the matrix unit with $1_R$ in the $(i,j)$-entry and $0_R$ elsewhere, and put $e:=e_{11}$. The assumption $n\ge 1$ is what permits us to choose this first coordinate. The corner $eAe$ consists exactly of matrices whose only possibly nonzero entry is in position $(1,1)$, so the map $R\to eAe$ that places $r\in R$ in the $(1,1)$-entry is a unital ring isomorphism onto the corner ring. Let $\operatorname{Mod}(A)$ denote the category of left $A$-modules and $A$-linear maps, and let $\operatorname{Mod}(R)$ denote the category of left $R$-modules and $R$-linear maps. For a left $A$-module $M$, define $F(M)=eM$. This is a left $R$-module through the identification $R\cong eAe$. Conversely, for a left $R$-module $N$, define $G(N)=Ae\otimes_R N$. The left $A$-module structure on $G(N)$ comes from left multiplication on the factor $Ae$. We now check why these functors are inverse. For a left $R$-module $N$, the module $F(G(N))$ is \begin{align*} e(Ae\otimes_R N). \end{align*} Since $eAe\cong R$, the map \begin{align*} e(Ae\otimes_R N)&\to N \end{align*} sending $e(ae\otimes x)$ to $eae\cdot x$ is well-defined and $R$-linear. Its inverse sends $x\in N$ to $e(e\otimes x)\in e(Ae\otimes_R N)$. Composing first from $N$ to $e(Ae\otimes_R N)$ and back gives \begin{align*} x \mapsto e(e\otimes x) \mapsto eee\cdot x = 1_R x = x. \end{align*} Composing in the other direction uses the tensor relation over $R\cong eAe$: the element $e(ae\otimes x)$ equals $e\otimes(eae\cdot x)$ in $e(Ae\otimes_R N)$, so the second composite is also the identity. Hence this comparison map is an isomorphism. For a left $A$-module $M$, the comparison map $\mu_M:Ae\otimes_R eM\to M$ sends $ae\otimes y$ to $ay$. This is $A$-linear. It is surjective because the matrix units satisfy \begin{align*} 1_A=\sum_{j=1}^{n} e_{j1}e_{1j}, \end{align*} so every $m\in M$ decomposes as a finite sum \begin{align*} m=\sum_{j=1}^{n} e_{j1}(e_{1j}m), \end{align*} and each $e_{1j}m$ lies in $eM$. Define $\nu_M:M\to Ae\otimes_R eM$ by $\nu_M(m)=\sum_{j=1}^{n}e_{j1}\otimes e_{1j}m$. The relation $e e_{1j}=e_{1j}$ shows that $e_{1j}m\in eM$, so $\nu_M$ is well-defined. The relation $e_{ab}e_{cd}=\delta_{bc}e_{ad}$ gives $\mu_M(\nu_M(m))=m$ for every $m\in M$, and the same relation gives $\nu_M(\mu_M(ae\otimes y))=ae\otimes y$ for every elementary tensor $ae\otimes y$. Hence $\mu_M$ is an isomorphism. Finally, finite projectivity is preserved because these functors preserve finite direct sums and direct summands, and because the images of the rank-one free generators are finitely generated projective. Indeed $F(A)=eA$ is isomorphic to $R^n$ as a left $R$-module by reading off the first row under $eAe\cong R$, while $G(R)=Ae$ is a direct summand of the finite free left $A$-module $A=Ae\oplus A(1_A-e)$. Therefore a direct summand of a finite free module is sent to a finitely generated projective module on both sides.[/guided]

[step:Pass from the projective-module monoids to $K_0$] Let $V(A)$ denote the commutative monoid of isomorphism classes of finitely generated projective left $A$-modules under direct sum, and similarly define $V(R)$. The functor $F$ induces a monoid homomorphism \begin{align*} V(A)&\to V(R) \end{align*} sending $[P]$ to $[eP]$. The inverse functor $G$ induces the inverse monoid homomorphism \begin{align*} V(R)&\to V(A) \end{align*} sending $[Q]$ to $[Ae\otimes_R Q]$. Hence $V(A)\cong V(R)$. The monoids $V(A)$ and $V(R)$ are commutative under direct sum. By the definition of $K_0$ as the group completion of $V(-)$, a monoid isomorphism $V(A)\cong V(R)$ induces an isomorphism of the corresponding group completions: apply the monoid isomorphism and its inverse, then use the universal property of group completion to obtain inverse group homomorphisms. Hence this monoid isomorphism induces a group isomorphism \begin{align*} K_0(M_n(R))\cong K_0(R). \end{align*} [/step]

[step:Expand stable invertible matrices over $M_n(R)$ into stable invertible matrices over $R$] For each $m\ge 1$, define the block expansion map $\beta_m:M_m(M_n(R))\to M_{mn}(R)$ by replacing each entry of an $m\times m$ matrix over $M_n(R)$ with its corresponding $n\times n$ matrix over $R$. This is a unital ring isomorphism, and therefore restricts to a group isomorphism $\beta_m:GL_m(M_n(R))\to GL_{mn}(R)$. These maps commute with stabilization. Under the inclusion $GL_m(M_n(R))\to GL_{m+1}(M_n(R))$ given by $B\mapsto \operatorname{diag}(B,1)$, block expansion gives \begin{align*} \beta_{m+1}(\operatorname{diag}(B,1))=\operatorname{diag}(\beta_m(B),I_n). \end{align*} Thus the stable group $GL(M_n(R))$ is identified with the direct limit of the cofinal subsystem \begin{align*} GL_n(R)\subset GL_{2n}(R)\subset GL_{3n}(R)\subset \cdots \end{align*} inside $GL(R)$. Since for every $k\ge 1$ there exists $m\ge 1$ with $k\le mn$, this subsystem is cofinal in the usual stabilization system for $GL(R)$. Therefore block expansion induces a group isomorphism \begin{align*} \beta:GL(M_n(R))&\to GL(R). \end{align*} [/step]

[step:Show that block expansion identifies the stable elementary subgroups] Let $E(M_n(R))\le GL(M_n(R))$ and $E(R)\le GL(R)$ denote the stable elementary subgroups. We prove \begin{align*} \beta(E(M_n(R)))=E(R). \end{align*} For a unital ring $B$, integers $q\ge 2$, distinct indices $i,j\in\{1,\dots,q\}$, and an element $b\in B$, write $\epsilon_{ij}(b):=I_q+bE_{ij}\in GL_q(B)$ for the elementary transvection, where $E_{ij}$ is the ordinary matrix unit in $M_q(B)$. First let $\epsilon_{ab}(X)$ be an elementary transvection in $GL_m(M_n(R))$, where $a\ne b$ and $X\in M_n(R)$. Write $X=(x_{\alpha\beta})_{\alpha,\beta=1}^{n}$. Its block expansion is \begin{align*} I_{mn}+\sum_{\alpha=1}^{n}\sum_{\beta=1}^{n} x_{\alpha\beta} E_{(a,\alpha),(b,\beta)} \end{align*} in $GL_{mn}(R)$, where $(a,\alpha)$ denotes the row index in the $a$-th block and internal row $\alpha$. Because $a\ne b$, the ordinary matrix units appearing in this sum multiply pairwise to $0$. Hence this matrix is the product of the ordinary elementary transvections \begin{align*} \prod_{\alpha=1}^{n}\prod_{\beta=1}^{n}\left(I_{mn}+x_{\alpha\beta}E_{(a,\alpha),(b,\beta)}\right). \end{align*} Thus $\beta(E(M_n(R)))\subset E(R)$. Conversely, let $I_k+rE_{uv}\in GL_k(R)$ be an ordinary elementary transvection, where $u\ne v$ and $r\in R$. Choose $m\ge 1$ such that $k\le mn$, and view this transvection in $GL_{mn}(R)$ by stabilization. Write the row and column indices as block indices \begin{align*} u=(a,\alpha),\qquad v=(b,\beta) \end{align*} with $1\le a,b\le m$ and $1\le \alpha,\beta\le n$. If $a\ne b$, then the transvection is the block expansion of the elementary transvection over $M_n(R)$ with block entry $r e_{\alpha\beta}$ in position $(a,b)$. If $a=b$, enlarge once more to $GL_{(m+1)n}(R)$ and choose a block index $c\ne a$. We use the commutator convention $[g,h]=ghg^{-1}h^{-1}$. The ring in question is the arbitrary unital ring $R$, and after stabilization the three coordinate positions $(a,\alpha)$, $(c,1)$, and $(a,\beta)$ are distinct when $\alpha\ne\beta$. The elementary matrix calculation for distinct positions $p,q,t$ is \begin{align*} [I+rE_{pq},I+E_{qt}]=I+rE_{pt}, \end{align*} because $E_{pq}E_{qt}=E_{pt}$ while the reverse product and the higher nilpotent cross terms vanish. Applying this relation with $p=(a,\alpha)$, $q=(c,1)$, and $t=(a,\beta)$ gives \begin{align*} I+rE_{(a,\alpha),(a,\beta)}=[I+rE_{(a,\alpha),(c,1)}, I+E_{(c,1),(a,\beta)}] \end{align*} when $\alpha\ne \beta$. Each factor on the right has distinct block indices, so each factor is the block expansion of an elementary transvection over $M_n(R)$. The case $\alpha=\beta$ cannot occur for an ordinary elementary transvection with $u\ne v$ when $a=b$, because then $u=(a,\alpha)=v$. Hence every ordinary elementary generator of $E(R)$ lies in $\beta(E(M_n(R)))$ after stabilization. Therefore $E(R)\subset\beta(E(M_n(R)))$, proving equality. [/step]

[step:Take quotients to obtain the $K_1$ isomorphism] By the convention for low algebraic $K$-theory used here, \begin{align*} K_1(A) := GL(A)/E(A). \end{align*} Here $E(A)\trianglelefteq GL(A)$ is the stable elementary subgroup generated by elementary transvections; this normality is part of the Whitehead presentation fixed in the theorem statement, so the displayed quotient is a group quotient. Applying the block expansion isomorphism $\beta:GL(M_n(R))\to GL(R)$ together with \begin{align*} \beta(E(M_n(R)))=E(R), \end{align*} we obtain an induced group isomorphism $K_1(M_n(R))\to K_1(R)$. This proves Morita invariance for $K_1$. [/step]

[step:Check naturality with respect to unital ring homomorphisms] Let $\varphi:R\to S$ be a unital ring homomorphism. It induces a unital ring homomorphism \begin{align*} M_n(\varphi):M_n(R)&\to M_n(S) \end{align*} by applying $\varphi$ to each matrix entry. For $K_0$, write $A_R:=M_n(R)$, $A_S:=M_n(S)$, $e_R:=e_{11}\in A_R$, and $e_S:=e_{11}\in A_S$. The homomorphism $M_n(\varphi)$ sends $e_R$ to $e_S$. If $P$ is a finitely generated projective left $A_R$-module, the canonical map \begin{align*} S\otimes_R e_RP &\to e_S(A_S\otimes_{A_R}P) \end{align*} sending $s\otimes e_Rp$ to $e_S((sE_{11})\otimes p)$ is an isomorphism of left $S$-modules, with inverse induced by writing an element of $e_S(A_S\otimes_{A_R}P)$ in the first row and using the corner identification $e_SA_Se_S\cong S$. This identifies applying the Morita functor after extension of scalars along $M_n(\varphi)$ with extending scalars along $\varphi$ after applying the Morita functor. Hence the square formed by \begin{align*} K_0(M_n(R))&\to K_0(R) \end{align*} and \begin{align*} K_0(M_n(S))&\to K_0(S) \end{align*} commutes. For $K_1$, the block expansion maps commute with applying $\varphi$ entrywise: \begin{align*} \beta_m(M_m(M_n(\varphi))(B))=M_{mn}(\varphi)(\beta_m(B)) \end{align*} for every $B\in GL_m(M_n(R))$. Passing to stable general linear groups and then to the quotients by elementary subgroups gives a commuting square on $K_1$. Therefore the isomorphisms $K_0(M_n(R))\cong K_0(R)$ and $K_1(M_n(R))\cong K_1(R)$ are natural in the unital ring $R$. This completes the proof. [/step]