[proofplan]
We relabel the entries of $t$ by the unique permutation $w\in S_n$ that carries each box entry of $t$ to the corresponding box entry of $s$. This relabeling conjugates the row and column stabilizers, and therefore conjugates the associated row symmetrizers, column antisymmetrizers, and Young symmetrizers. The identity $c_s=wc_tw^{-1}$ rewrites $\mathbb C[S_n]c_s$ as $\mathbb C[S_n]c_tw^{-1}$, and right multiplication by $w$ gives the desired left-module isomorphism because right multiplication commutes with the left regular action.
[/proofplan]
[step:Choose the permutation relabelling $t$ to $s$]
Let $D_\lambda$ denote the Young diagram of $\lambda$. Since $s$ and $t$ are bijective fillings of $D_\lambda$ by $\{1,\dots,n\}$, there is a unique permutation $w\in S_n$ such that, for every box $B\in D_\lambda$,
\begin{align*}
s(B)=w(t(B)).
\end{align*}
Equivalently, the natural relabelling action of $S_n$ on tableaux satisfies $s=wt$.
[/step]
[step:Conjugate the row and column stabilizers]
We claim that
\begin{align*}
R_s=wR_tw^{-1}
\end{align*}
and
\begin{align*}
C_s=wC_tw^{-1}.
\end{align*}
Indeed, a permutation $\rho\in S_n$ lies in $R_t$ precisely when it preserves each row-set of entries of $t$. Since $w$ sends every row-set of entries of $t$ to the corresponding row-set of entries of $s$, the permutation $w\rho w^{-1}$ preserves each row-set of entries of $s$. This proves $wR_tw^{-1}\subset R_s$. Applying the same argument to $w^{-1}$ gives the reverse inclusion. For columns, a permutation $\gamma\in S_n$ lies in $C_t$ precisely when it preserves each column-set of entries of $t$. Since $w$ sends every column-set of entries of $t$ to the corresponding column-set of entries of $s$, the permutation $w\gamma w^{-1}$ preserves each column-set of entries of $s$, so $wC_tw^{-1}\subset C_s$. Applying this column-set argument to $w^{-1}$ gives $C_s\subset wC_tw^{-1}$.
[guided]
The point of introducing $w$ is that it transports the combinatorics of $t$ box by box to the combinatorics of $s$. Let $B\in D_\lambda$ be a box. The defining relation $s(B)=w(t(B))$ says that the set of entries appearing in any row of $s$ is the image under $w$ of the set of entries appearing in the corresponding row of $t$.
We prove the row-stabilizer identity first. A permutation $\rho\in S_n$ belongs to $R_t$ exactly when, for every row of $D_\lambda$, the set of labels in that row of $t$ is mapped to itself by $\rho$. Conjugating by $w$ first translates an $s$-label back to a $t$-label, applies $\rho$, and then translates forward again. Therefore $w\rho w^{-1}$ preserves every row-set of $s$, so $w\rho w^{-1}\in R_s$. This gives
\begin{align*}
wR_tw^{-1}\subset R_s.
\end{align*}
For the opposite inclusion, take $\sigma\in R_s$. Since $w^{-1}$ sends each row-set of $s$ back to the corresponding row-set of $t$, the conjugate $w^{-1}\sigma w$ preserves every row-set of $t$. Hence $w^{-1}\sigma w\in R_t$, and therefore $\sigma\in wR_tw^{-1}$. Thus
\begin{align*}
R_s=wR_tw^{-1}.
\end{align*}
The same argument applies to columns. A permutation lies in $C_t$ exactly when it preserves each column-set of entries of $t$, and $w$ sends the column-sets of $t$ to the corresponding column-sets of $s$. Hence
\begin{align*}
C_s=wC_tw^{-1}.
\end{align*}
[/guided]
[/step]
[step:Transport the Young symmetrizer by conjugation]
Using the row-stabilizer identity and the substitution $\rho=w\rho_t w^{-1}$ with $\rho_t\in R_t$, we get
\begin{align*}
a_s=\sum_{\rho\in R_s}\rho=\sum_{\rho_t\in R_t}w\rho_t w^{-1}=wa_tw^{-1}.
\end{align*}
Similarly, since the sign character is invariant under conjugation,
\begin{align*}
\operatorname{sgn}(w\gamma_t w^{-1})=\operatorname{sgn}(\gamma_t)
\end{align*}
for every $\gamma_t\in C_t$, and therefore
\begin{align*}
b_s=\sum_{\gamma\in C_s}\operatorname{sgn}(\gamma)\gamma=\sum_{\gamma_t\in C_t}\operatorname{sgn}(\gamma_t)w\gamma_t w^{-1}=wb_tw^{-1}.
\end{align*}
With the convention $c_u=a_ub_u$, this gives
\begin{align*}
c_s=a_sb_s=(wa_tw^{-1})(wb_tw^{-1})=wa_tb_tw^{-1}=wc_tw^{-1}.
\end{align*}
[/step]
[step:Rewrite the left ideal generated by $c_s$]
From $c_s=wc_tw^{-1}$, we have
\begin{align*}
\mathbb C[S_n]c_s=\mathbb C[S_n]wc_tw^{-1}.
\end{align*}
Since $w\in S_n$ is a unit of the group algebra, left multiplication by $w$ maps $\mathbb C[S_n]$ bijectively onto $\mathbb C[S_n]$. Hence
\begin{align*}
\mathbb C[S_n]w=\mathbb C[S_n].
\end{align*}
Substituting this into the previous equality gives
\begin{align*}
\mathbb C[S_n]c_s=\mathbb C[S_n]c_tw^{-1}.
\end{align*}
[/step]
[step:Use right multiplication by $w$ to build the module isomorphism]
Define a map
\begin{align*}
\Phi:\mathbb C[S_n]c_s &\to \mathbb C[S_n]c_t
\end{align*}
by
\begin{align*}
\Phi(x)=xw
\end{align*}
for $x\in\mathbb C[S_n]c_s=\mathbb C[S_n]c_tw^{-1}$. If $x=yc_tw^{-1}$ with $y\in\mathbb C[S_n]$, then
\begin{align*}
\Phi(x)=yc_t\in\mathbb C[S_n]c_t,
\end{align*}
so $\Phi$ is well-defined as a map into $\mathbb C[S_n]c_t$.
The inverse map is right multiplication by $w^{-1}$:
\begin{align*}
\Psi:\mathbb C[S_n]c_t &\to \mathbb C[S_n]c_s
\end{align*}
with
\begin{align*}
\Psi(z)=zw^{-1}.
\end{align*}
Let $e\in S_n$ denote the identity permutation. Indeed, for $z=yc_t$ we have $zw^{-1}=yc_tw^{-1}\in\mathbb C[S_n]c_s$, and the identities $\Phi(\Psi(z))=z$ and $\Psi(\Phi(x))=x$ follow from $ww^{-1}=w^{-1}w=e$.
Finally, $\Phi$ is $\mathbb C[S_n]$-linear for the left module structures. For every $g\in S_n$ and every $x\in\mathbb C[S_n]c_s$,
\begin{align*}
\Phi(gx)=(gx)w=g(xw)=g\Phi(x).
\end{align*}
By complex linearity, the same identity holds for every element of $\mathbb C[S_n]$ in place of $g$. Thus $\Phi$ is an isomorphism of left $\mathbb C[S_n]$-modules.
[/step]
[step:Deduce independence of the chosen tableau]
Since $s$ and $t$ were arbitrary bijective Young tableaux of shape $\lambda$, the left ideals $\mathbb C[S_n]c_s$ and $\mathbb C[S_n]c_t$ are isomorphic for any two choices of tableau. Therefore the isomorphism class of the Specht module
\begin{align*}
S^\lambda:=\mathbb C[S_n]c_t
\end{align*}
depends only on the partition $\lambda$, not on the chosen tableau $t$ of shape $\lambda$.
[/step]
