Allen's nice answer led me in a slightly different direction. Let me try to give another answer to the question.

Let's start from the Cauchy identities, $$ \prod_{i \geq 0} (1-x_i y_i)^{-1} = \sum_\lambda s_\lambda(x)s_\lambda(y) $$ which is an equality between bisymmetric functions in infinitely many variables $\{x_i\}$ and $\{y_i\}$.

Now recall that there is a correspondence between symmetric functions of degree $n$, representations of $S_n$, and polynomial functors $\mathrm{Vect}_{\mathbb C}\to\mathrm{Vect}_{\mathbb C}$ of degree $n$. Passing to the completion of the ring of symmetric functions wrt degree gives instead a correspondence between (completed) symmetric functions, sequences of representations of $S_n$ (i.e. "tensorial species") and analytic functors $\mathrm{Vect}_{\mathbb C}\to\mathrm{Vect}_{\mathbb C}$.

Using this we can give three different interpretations of the Cauchy identities:

(1) Consider both the $x$- and $y$-variables as corresponding to representations of the symmetric groups. The Cauchy identities become $$ \bigoplus_{n \geq 0} \mathbb C[S_n] = \bigoplus_{\lambda} \sigma_\lambda \otimes \sigma_\lambda,$$ i.e. the Peter-Weyl theorem for $S_n$.

(2) Consider the $x$-variables as corresponding to an analytic functor and the $y$-variables as corresponding to a sequence of representations. Then the left hand side becomes the analytic functor $V \mapsto T(V)$ and the right hand side becomes $V \mapsto \bigoplus_\lambda V_\lambda \otimes \sigma_\lambda$.

(3) Consider both $x$- and $y$-variables as corresponding to analytic functors. The left hand side becomes the analytic functor $(V,W) \mapsto \mathcal O(V\otimes W)$ and the right hand side becomes $(V,W) \mapsto \bigoplus_\lambda V_\lambda \otimes W_\lambda$.

Specializing to $W = V^\ast$ in (3) gives the coordinate ring of the matrix space as in Allen's answer.

The three interpretations of the Cauchy identities can be seen as equalities between sequences of representations of $S_n \times S_n$, sequences of polynomial functors of degree $n$ into $S_n$-representations, and analytic functors of two variables, respectively. But in all cases there is also an obvious multiplication on the left hand side: given by the inclusion $\mathbb C[S_n] \otimes \mathbb C[S_m] \to \mathbb C[S_{n+m}]$, the multiplication in the tensor algebra, and the multiplication in the coordinate ring, respectively. This is because in all three cases we have a commutative algebra object in the respective symmetric monoidal category, and the structure of commutative algebra object gets transferred along the different equivalences of categories. For example, a commutative algebra object in the category of tensorial species is what's usually called a twisted commutative algebra, so we are saying that the tensor algebra $T(V)$ is a twisted commutative algebra (even though the multiplication in $T(V)$ is certainly not commutative), and so on. So the multiplication in $T(V)$ is in a precise sense "the same" as the multiplication in the coordinate ring of $V \otimes W$!

PS - I certainly hope all the above is correct. But I am confused about the fact that what appears is $\sigma_\lambda \otimes \sigma_\lambda$ in case (1), rather than $\sigma_\lambda \otimes \sigma_\lambda^\ast$ which would be more expected. (Of course $\sigma_\lambda \cong \sigma_\lambda^\ast$, but I would still like the dual to be there!)

Allen's nice answer led me in a slightly different direction. Let me try to give another answer to the question.

Let's start from the Cauchy identities, $$ \prod_{i,j \geq 0} (1-x_i y_j)^{-1} = \sum_\lambda s_\lambda(x)s_\lambda(y) $$ which is an equality between bisymmetric functions in infinitely many variables $\{x_i\}$ and $\{y_i\}$.

Now recall that there is a correspondence between symmetric functions of degree $n$, representations of $S_n$, and polynomial functors $\mathrm{Vect}_{\mathbb C}\to\mathrm{Vect}_{\mathbb C}$ of degree $n$. Passing to the completion of the ring of symmetric functions wrt degree gives instead a correspondence between (completed) symmetric functions, sequences of representations of $S_n$ (i.e. "tensorial species") and analytic functors $\mathrm{Vect}_{\mathbb C}\to\mathrm{Vect}_{\mathbb C}$.

Using this we can give three different interpretations of the Cauchy identities:

(1) Consider both the $x$- and $y$-variables as corresponding to representations of the symmetric groups. The Cauchy identities become $$ \bigoplus_{n \geq 0} \mathbb C[S_n] = \bigoplus_{\lambda} \sigma_\lambda \otimes \sigma_\lambda,$$ i.e. the Peter-Weyl theorem for $S_n$.

(2) Consider the $x$-variables as corresponding to an analytic functor and the $y$-variables as corresponding to a sequence of representations. Then the left hand side becomes the analytic functor $V \mapsto T(V)$ and the right hand side becomes $V \mapsto \bigoplus_\lambda V_\lambda \otimes \sigma_\lambda$.

(3) Consider both $x$- and $y$-variables as corresponding to analytic functors. The left hand side becomes the analytic functor $(V,W) \mapsto \mathcal O(V\otimes W)$ and the right hand side becomes $(V,W) \mapsto \bigoplus_\lambda V_\lambda \otimes W_\lambda$.

Specializing to $W = V^\ast$ in (3) gives the coordinate ring of the matrix space as in Allen's answer.

The three interpretations of the Cauchy identities can be seen as equalities between sequences of representations of $S_n \times S_n$, sequences of polynomial functors of degree $n$ into $S_n$-representations, and analytic functors of two variables, respectively. But in all cases there is also an obvious multiplication on the left hand side: given by the inclusion $\mathbb C[S_n] \otimes \mathbb C[S_m] \to \mathbb C[S_{n+m}]$, the multiplication in the tensor algebra, and the multiplication in the coordinate ring, respectively. This is because in all three cases we have a commutative algebra object in the respective symmetric monoidal category, and the structure of commutative algebra object gets transferred along the different equivalences of categories. For example, a commutative algebra object in the category of tensorial species is what's usually called a twisted commutative algebra, so we are saying that the tensor algebra $T(V)$ is a twisted commutative algebra (even though the multiplication in $T(V)$ is certainly not commutative), and so on. So the multiplication in $T(V)$ is in a precise sense "the same" as the multiplication in the coordinate ring of $V \otimes W$!

PS - I certainly hope all the above is correct. But I am confused about the fact that what appears is $\sigma_\lambda \otimes \sigma_\lambda$ in case (1), rather than $\sigma_\lambda \otimes \sigma_\lambda^\ast$ which would be more expected. (Of course $\sigma_\lambda \cong \sigma_\lambda^\ast$, but I would still like the dual to be there!)