As I'm sure you know, this is an easy computation with symmetric polynomials/characters over the field $\mathbb{C}$. I will give a proof which shows how to leverage that fact and show that, for any commutative $\mathbb{Q}$-algebra $R$, and $U$ and $V$ any free $R$-modules, we have the required isomorphism. At the end, I'll try to make a general statement.

**Lemma 1:** Let $U$ a finite dimensional $\mathbb{C}$ vector space. For $\sigma$ a permutation in $S_d$, let $\sigma : U^{\otimes d} \to U^{\otimes d}$ act by permuting the factors according to $\sigma$. So we get a map
$$\mathbb{C}[S_d] \to \mathrm{End}_{GL(U)}(U^{\otimes d}).$$
This map is surjective.

**Proof:** This is part of the statement of Schur-Weyl duality.

**Corollary 2:** Let $U$ and $V$ be finite dimensional $\mathbb{C}$ vector spaces. We have similar surjections:
$$\mathbb{C}[S_d \times S_e] \to \mathrm{End}_{GL(U) \times GL(V)}(U^{\otimes d} \otimes V^{\otimes e})$$
and
$$\mathrm{Mat}_{r \times s}( \mathbb{C}[S_e \times S_d]) \to \mathrm{Hom}_{GL(U) \times GL(V)} \left( \left(U^{\otimes d} \otimes V^{\otimes e}\right)^{\oplus r}, \left(U^{\otimes d} \otimes V^{\otimes e}\right)^{\oplus s} \right).$$

Now, let $U$ and $V$ be finite dimensional $\mathbb{Q}$-vector spaces. Observe that

$$\bigwedge\nolimits^{2}\left(U\otimes V\right)\oplus \left(\bigwedge\nolimits^2\left(U\right)\otimes\bigwedge\nolimits^2\left(V\right)\right)^{\oplus 2} \quad (\ast)$$
is naturally a quotient of $(U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 3}$. Even better, there is an idempotent $\pi_1$ in $\mathrm{Mat}_{3 \times 3}(\mathbb{Q}[S_2 \times S_2]$ such that $(\ast)$ is the image of $\pi_1$.

Similarly, we can find $\pi_2$, an idempotent in $\mathrm{Mat}_{2 \times 2}(\mathbb{Q}[S_d \times S_e])$ such that the image of $\pi_2$ acting on $(U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 2}$ is
$$U^{\otimes 2}\otimes\bigwedge\nolimits^2 V \oplus \bigwedge\nolimits^2 U\otimes V^{\otimes 2} \quad (\ast\ast)$$

We have a commutative diagram
$$\begin{matrix}
\mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_2 \times S_2]) & \rightarrow & \mathrm{Mat}_{2 \times 3}(\mathbb{C}[S_2 \times S_2]) \\
\downarrow & & \downarrow \\
\mathrm{Hom}_{GL(U) \times GL(V)} \left( (U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 3}, (U^{\otimes 2} \otimes V^{\otimes 2})^{\oplus 2} \right) & \rightarrow & \mathrm{Hom}_{GL(U_{\mathbb{C}}) \times GL(V_{\mathbb{C}})} \left( (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 3}, (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 2} \right)
\end{matrix}$$
where for brevity I use a subscript $\mathbb{C}$ for tensor with $\mathbb{C}$. Call the vertical maps $\alpha$.

Let $A \subset \mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_d])$ be those maps $\phi$ such that
$$\alpha(\phi) = \alpha(\phi \pi_1) = \alpha(\pi_2 \phi). \quad (\dagger)$$
Let $A_{\mathbb{C}}$ be the analogous subspace of $\mathrm{Mat}_{2 \times 3}(\mathbb{C}[S_d])$. Since $A$ and $A_{\mathbb{C}}$ are cut out by the same equations, we have $A_{\mathbb{C}} = A \otimes \mathbb{C}$.

The images of $\pi_1$ and $\pi_2$ are isomorphic over $\mathbb{C}$. Composing that isomorphism with the projection $(U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 3} \to (\ast)_{\mathbb{C}}$ and the injection $(\ast \ast)_{\mathbb{C}} \to (U_{\mathbb{C}}^{\otimes 2} \otimes V_{\mathbb{C}}^{\otimes 2})^{\oplus 2}$, we get a map $\overline{\phi}$ in the bottom right hand corner of the diagram, obeying $\overline{\phi} = \overline{\phi} \alpha(\pi_1) = \alpha(\pi_1) \overline{\phi}$. Moreover, this map has rank $\dim (\ast) = \dim (\ast \ast)$.

Using Corollary 2, we can find $\phi$ with $\alpha(\phi) = \overline{\phi}$. So $\phi \in A_{\mathbb{C}}$ and $\alpha(\phi)$ has rank $\dim (\ast)$.

**Crucial paragraph** The subspace of $A_{\mathbb{C}}$ consisting of maps $\phi$ such that $\alpha(\phi)$ has rank $\dim (\ast)$ is Zariski open. And $A$ is Zariski dense in $A_{\mathbb{C}}$. (Any polynomial which vanishes at all rational inputs is zero.) So there is some $\phi \in A$ obeying $(\dagger)$. This $\phi$ must induce an isomorphism between $(\ast)$ and $(\ast \ast)$.

So we have shown that $(\ast) \cong (\ast \ast)$, and this isomorphism is induced by a matrix in $\mathrm{Mat}_{2 \times 3}(\mathbb{Q}[S_d])$. For any $\mathbb{Q}$ algebra $R$, this gives a corresponding isomorphism between the analogous representations of $GL(U \otimes R) \times GL(V \otimes R)$.

Wow, that took a lot longer than I expected it to. I dread the notation that would be needed to make this precise, but a theorem like the following should follow by similar arguments: Let $\mathcal{S}$ and $\mathcal{T}$ be two formulas made up from atoms $V_1$, $V_2$ ..., $V_r$ joined together direct sums, tensor products and Schur functors. For any commutative ring $R$ and sequence of positive integers $(n_1, \ldots, n_r)$, we can "plug in" $R^{n_i}$ for $V_i$ and get a $\prod GL_{n_i}(R)$ representation.

**"Theorem"** If $\mathcal{S}(\mathbb{C}, n_1, \ldots, n_r) \cong \mathcal{T}(\mathbb{C}, n_1, \ldots, n_r)$, then $\mathcal{S}(R, n_1, \ldots, n_r) \cong \mathcal{T}(R, n_1, \ldots, n_r)$ holds for any $\mathbb{Q}$-algebra $R$ and there is a "universal formula" for the isomorphism as a $\mathbb{Q}$-linear combination of reordering various tensor products.

When I set out to prove this, I expected to only need that all positive integers $\leq \max(n_1, \ldots, n_r)$ were invertible in $R$, not that $R$ was a $\mathbb{Q}$ algebra. But I couldn't get the Zariski density argument in the Crucial Paragraph to work without a ground field. I'm curious whether this was an artifact of the proof method, or a genuine obstacle.