Let's assume that all the integers are in fact at least $1$.
Let $\max(P) = \prod_i s_i$. If $\max(P) \neq \min(P)$ then $\max(P)-1$ must be representable. Now $(s_1 - 1) \prod_{i>1} s_i = \prod_i s_i - \prod_{i>1} s_i$, so there must be an index $I$ such that $S_i = \{1\}$ for $i \neq I$. $S_I$ itself is of course of the form $\{1,\ldots,\max(P)\}$.
Summarizing, we have found two solutions: $|S_i|=1$ for all $i$ (the sets are otherwise arbitrary), and $S_i = \{1\}$ apart for one set which is an interval.
If the sets are infinite (i.e. $\max(P) = \infty$) then for each prime $p$ take the set $S_p = \{p^k : k \geq 0\}$. There are other solutions: you can decompose the set of primes arbitrarily as the disjoint union $\bigcup P_i$ and then take as $S_i$ all numbers whose prime factors are in $P_i$. All these use the convention that you only select finitely many numbers which are not $1$.
Open question: are these the only solutions?