[proofplan]
The proof reduces the norm of an arbitrary element $a$ to the spectral radius of the positive element $a^*a$. For each of the two C*-norms, the C*-identity gives $\|a\|_j^2=\|a^*a\|_j$, and the standard spectral-radius formula in C*-algebras identifies the norm of the self-adjoint element $a^*a$ with its algebraic spectral radius. That spectral radius is computed in the same algebraic unitization, so it is independent of which C*-norm was used. Taking square roots then gives equality of the two norms.
[/proofplan]
[step:Place both norms in the same algebraic unitization]
Let $A^+$ denote the algebraic unitization of $A$. Thus
\begin{align*}A^+ := A\oplus \mathbb C 1,\end{align*}
with multiplication determined by
\begin{align*}
(a+\lambda 1)(b+\mu 1)=ab+\lambda b+\mu a+\lambda\mu 1
\end{align*}
and involution determined by
\begin{align*}(a+\lambda 1)^*=a^*+\overline{\lambda}1\end{align*}
for all $a,b\in A$ and all $\lambda,\mu\in\mathbb C$.
If $A$ is already unital, this notation means the common algebraic unit of $A$. If $A$ is nonunital, this is the usual adjoined unit. In either case, for each $x\in A^+$ the spectrum
\begin{align*}\sigma_{A^+}(x):=\{\lambda\in\mathbb C:x-\lambda 1\text{ is not invertible in }A^+\}\end{align*}
depends only on the multiplication and unit of $A^+$, not on either norm. Define the corresponding spectral radius by
\begin{align*}
r_{A^+}(x):=\sup\{|\lambda|:\lambda\in\sigma_{A^+}(x)\}.
\end{align*}
[/step]
[step:Identify the norm of a self-adjoint element with its algebraic spectral radius]
We use the standard C*-algebra spectral-radius fact: if $B$ is a C*-algebra and $x=x^*\in B$, then
\begin{align*}
\|x\|_B=r_{B^+}(x),
\end{align*}
where the spectral radius is computed in the unitization when $B$ is nonunital.
Apply this to the C*-algebra $(A,\|\cdot\|_j)$ for each $j\in\{1,2\}$. Since the multiplication and involution are the same for both C*-algebra structures, the element $a^*a$ is self-adjoint in both:
\begin{align*}(a^*a)^*=a^*a.\end{align*}
Therefore, for each $j\in\{1,2\}$,
\begin{align*}
\|a^*a\|_j=r_{A^+}(a^*a).
\end{align*}
[guided]
We now isolate the only analytic input in the proof. In a C*-algebra, the norm of a self-adjoint element is determined by its spectrum: if $B$ is a C*-algebra and $x=x^*\in B$, then
\begin{align*}
\|x\|_B=r_{B^+}(x).
\end{align*}
For a nonunital algebra, the notation $B^+$ means the algebraic unitization, because invertibility and spectrum require a unit.
We apply this result twice, once to the C*-algebra obtained from $A$ using $\|\cdot\|_1$ and once to the C*-algebra obtained from $A$ using $\|\cdot\|_2$. The element we feed into the result is
\begin{align*}
x:=a^*a.
\end{align*}
It is self-adjoint because the involution is anti-multiplicative:
\begin{align*}
x^*=(a^*a)^*=a^*a=x.
\end{align*}
Hence the spectral-radius formula gives
\begin{align*}
\|a^*a\|_1=r_{A^+}(a^*a)
\end{align*}
and
\begin{align*}
\|a^*a\|_2=r_{A^+}(a^*a).
\end{align*}
The key point is that the right-hand side is the same in both equations. The spectrum of $a^*a$ is defined by algebraic invertibility of $a^*a-\lambda 1$ in the common unitization $A^+$, and this invertibility relation does not mention either norm.
[/guided]
[/step]
[step:Use the C*-identity to compare the two norms of an arbitrary element]
Fix $a\in A$. For each $j\in\{1,2\}$, the C*-identity in the C*-algebra $(A,\|\cdot\|_j)$ gives
\begin{align*}
\|a\|_j^2=\|a^*a\|_j.
\end{align*}
By the preceding step,
\begin{align*}\|a^*a\|_1=r_{A^+}(a^*a)=\|a^*a\|_2.\end{align*}
Consequently,
\begin{align*}\|a\|_1^2=\|a\|_2^2.\end{align*}
Both norms take values in $[0,\infty)$, so taking the nonnegative square root gives
\begin{align*}\|a\|_1=\|a\|_2.\end{align*}
Since $a\in A$ was arbitrary, the two C*-norms agree on all of $A$.
[/step]
