Yes. In combinatorics this is known as Robinson-Schensted vs. Robinson-Schensted-Knuth.
First, shrink the Peter-Weyl result from $\mathcal O(GL(n))$ (you overuse $V$, I feel) to the slightly smaller $\mathcal O(M_n)$. Then the RHS shrinks to $\oplus_\lambda V_\lambda \otimes V_\lambda^*$, where $\lambda$ now runs over partitions $(\lambda_1 \geq \ldots \geq \lambda_n \geq 0)$ instead of all dominant weights $(\lambda_1 \geq \ldots \geq \lambda_n)$.
Then generalize to other matrix spaces, not just square matrices, obtaining $\mathcal O(M_{a\times b}) \cong \bigoplus_\lambda V^a_\lambda \otimes (V^b_\lambda)^*$, the sum now over partitions of height $\leq \min(a,b)$.
(The combinatorial statement, RSK, is a bijective proof of two different character formulae for this representation. The obvious weight basis is given by monomials in the matrix entries, equivalently listed as $M_{a\times b}(\mathbb N)$. On the RHS we have pairs of same-shape SSYT. Under the bijection the row and column sums of the matrix in $M_{a\times b}(\mathbb N)$ go to the contents, i.e. entry multiplicities, of the two SSYT.)
Now, consider functions on $M_{a\times b}$ of weight $(1,1,\ldots,1)$ under the $T^a \leq GL(a)$ action. Since that's $S_a$-invariant and $S_a$ normalizes $T^a$, this weight space will have a $S_a \times GL(b)$ action.
The LHS will be made of functions that are multilinear in the rows, i.e. $(\mathbb C^b)^{\otimes a}$. The representation $V^a_\lambda$ has a $(1,1,\ldots,1)$ weight space iff $\lambda$ is a partition of $a$, and in that case, the $S_a$ action on it is the Specht irrep $\sigma_\lambda$ of $S_a$. Which is to say, the RHS has become $\oplus_{\lambda \vdash n} \sigma_\lambda \otimes (V_\lambda)^*$ like you wanted. QED.
(Now we're insisting that the row sums are all $1$. On the RHS, one of the SSYT is an SYT. If you go further and ask that the column sums be all $1$ also, then the LHS becomes just permutation matrices, the RHS pairs of same-shape SYT, and the correspondence is just Robinson-Schensted no Knuth.)
As I recently learned from Martin Kassabov, you can run this in reverse: take two copies of the Schur-Weyl isomorphism, reverse one, and tensor them together over $\mathbb C[S_n]$ to get the Peter-Weyl (for matrices) result. So it's a matter of taste deciding which one is the more fundamental.