Question

I believe the answer to your question is yes.

The Reynolds operator $$ R: k[X_1,\ldots,X_n,Y_1,\ldots,Y_n] \longrightarrow k[X_1,\ldots,X_n,Y_1,\ldots,Y_n]^{S_n} $$ is $k[X_1,\ldots,X_n,Y_1,\ldots,Y_n]^{S_n}$-equivariant, and therefore in particular $k[X_1,\ldots,X_n,Y_1,\ldots,Y_n]^{S_n\times S_n}$-equivariant. Also, $R$ is a projection, and therefore the image of $R$ is a direct summand of of its domain of definition. It follows therefore that the $k[X_1,\ldots,X_n,Y_1,\ldots,Y_n]^{S_n\times S_n}$-module $ k[X_1,\ldots,X_n,Y_1,\ldots,Y_n]^{S_n}$ is a direct summand of the free $k[X_1,\ldots,X_n,Y_1,\ldots,Y_n]^{S_n\times S_n}$-module $k[X_1,\ldots,X_n,Y_1,\ldots,Y_n]$.

Now since the action of $S_n\times S_n$ is generated by reflections, its invariant ring is a polynomial ring, and therefore any direct summand of a free module over this ring is again free (this is the Quillen–Suslin theorem).

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Edit (concerning the rank, which is $n!$): Let $G$ be a finite group acting on $R=k[x_1,\ldots,x_n]$ and let $H\leq G$ be a subgroup. Then $$ {\rm frac}(R^G)\otimes_{R^G} R^H \cong (R^G-{0})^{-1}R^H \cong {\rm frac}(R^H) $$
where ${\rm frac}$ denotes the fraction field (this is easily seen from the fact that by a construction similar to the Reynols operator, the denominator of a fraction over $R^H$ can always be made $R^G$-invariant). Let $Q={\rm frac}(R)$, then one can show in a similar fashion $Q^G = {\rm frac}(R^G)$. We conclude that $$ {\rm rank}_{R^G} R^H = {\rm dim}_{{\rm frac}(R^G)} {\rm frac} (R^H) = [Q^H:Q^G] = [G:H] $$ the last equality being due to Galois theory.