Well, it was extremely messy, but I solved it. Such an $S'$ (which I call a perfect subset) always exists. I wrote up the proof here: https://isaacg1.github.io/assets/divisors.pdf.

The basic structure of the proof is a strong induction on $n$, the size of the multiset. I then did a case analysis on the factorization of $n$, and the amounts of different kinds of elements of $S$. When $S$ has many elements that are multiples of a given prime factor of $n$, it is possible to construct a perfect subset. When $S$ has many 1s, it is possible to construct a perfect subset. When $S$ has many elements greater than 1 that are not multiples of a given prime factor of $n$, it is possible to construct a perfect subset. Using these three scenarios repeatedly to eliminate various cases, I was eventually able to cover every possibility, showing that a perfect subset always exists.