This answer is partial. 

We assume that all $a_i$ are non-negative. Put $A=\tfrac 1m \sum a_i$ and $A_j=\sum_{i\in S_j} a_i$. We want to maximize $\Pi=\prod A_j$. But by AM-GM inequality, $P^{1/m}\le \tfrac 1m \sum A_j=A$  and the equality holds iff each $A_j$ equals $A$. 

Even when $m=2$ and all $a_i$ are positive integers the problem to check whether the equality can hold is a [variant](https://en.wikipedia.org/wiki/Partition_problem#Variants_and_generalizations) of the [partition problem](https://en.wikipedia.org/wiki/Partition_problem), and at the first reference is stated that it is $NP$-hard (but without a citation). So this should be a known problem and there can be already developed heuristic and approximation algorithms for it. 

I guess that a heuristics can be based on some balancing idea, which looks promising when there are no big gaps between $a_i$’s.

For instance, I can propose the following algorithm to find an initial feasible partition. Assign $m$ piles, order $a_i$ in a non-increasing order, and then split this sequence of $a_i$’s into bags $B_1,\dots, B_n$ each containing $m$ consecutive $a_i$’s. An each step we pick a next bag and distribute its $a_i$’s into the piles, one number to each pile, trying to make the vector of  the sums of elements of the piles more balanced. A measure of a balancedness of a vector $x=(x_1,\dots,x_m)$ can be a norm (for instance, $\ell_2$ or $\ell_\infty$) of a vector $x-\left(\frac 1m \sum x_i\right) (1,1,\dots,1)$.

The obtained feasible partition (or even a random one) can be further iterative balanced by local search. Namely, given sets $S_1,\dots S_m$ and a small constant $b\ge 1$ (maybe even $b=1$ will provide a good approximation) we check all subsets $C_i$ of size $b$ in each of $S_j$ (so there are  ${n\choose b}^m$ possibilities to consider in total). For each of the possibilities we consider a union $C$ of $C_i$ and try to redistribute $C$ between $S_j$ trying to make the sequence of their sums more balanced. In particular, when $b=1$ and $m=2$, we look for indices $i_1\in S_1$ and $i_2\in S_2$ such that when we swap $i_1$ and $i_2$ between $S_1$ and $S_2$, the difference $|A_1-A_2|$ will decrease.