Throughout, the index $i$ runs over $i=1,2$.
Let $A_i$ be disjoint sets with union $A$. Consider the symmetric groups $S^{(i)} = \mathrm{Symm}(A_i)$ and $S = \mathrm{Symm}(A)$. The $S^{(i)}$ embed as subgroups of $S$ in an obvious way. They have trivial intersection and commute with each other, and their product set is a proper subgroup of $G$. Is there some kind of standard binary operation, maybe call it , that takes two groups and forms a third, such that $S^{(1)} S^{(2)} \simeq S$?
Edit: to clarify, there is a standard term for the relationship between $S$ and the elementwise product (further edit here) of the embedded subgroups $T = S^{(1)} S^{(2)}$, which is in this case also the direct product $S^{(1)} \times S^{(2)}$. Specifically, $T$ is the stabilizer of the partition $A = A_1 \sqcup A_2$, and is a normal subgroup of $S$. The quotient $S/T$ measures the extent to which permutations of $A$ can mix up $A_1$ and $A_2$.