I think one can at least give a First Fundamental Theorem in the spirit of really classical invariant theory.
For the group $S_n$, I think you can build all invariants of vectors $(X_i)_{1\le i\le n}$ under the action by permutation matrices using tensor contractions with the following fundamental tensors $T^{(1)},\ldots,T^{(n)}$ given by:
$$
(T^{(k)})_{i_1,\ldots,i_k}=\prod_{1\le \alpha<\beta\le k} \delta_{i_{\alpha},i_{\beta}}\ .
$$
For instance, using Einstein's convention for summation over repeated indices,
$$
p_k=(T^{(k)})_{i_1,\ldots,i_k} X_{i_1}\ldots X_{i_k}
$$
give the power sums.
I think one can prove this following the very general recipe in this MO answer.
Now to go to a product $S_n\times S_n$ you can follow the procedure in this other MO answer.
Basically you need a duplicate collection of fundamental tensors, say $S^{(1)},\ldots,S^{(n)}$ given by the same formulas. Now if you have a product of $X_{ij}$ the rule is the $T$'s only contract to $i$-type row indices,
whereas the $S$'s only contract to the $j$-type column indices. Of course you need to contract everything in sight. This will give an infinite linearly generating set for the ring of invariants you are looking for indexed by bipartite graphs. The degree of an invariant is the number of edges. Now finding a finite subset of such graphs which will give a system of ring generators, that's the big problem I think Dima is referring to.
Some references which discuss the invariant theory for $S_n$ with the $T^{(k)}$ tensors are:
Pouzet's conjecture was that certain set $P'$ is generating the same ring as $P$, but this turned out to be false.
– Dima Pasechnik Nov 19 '15 at 23:26