Prove that a set $S$ cannot be partitioned into two subsets, each having the same products of elements

Let $p\equiv 3\mod 4$ be a prime. Let $S$ be a set of $p-1$ consecutive positive integers. Prove that $S$ cannot be partitioned into two subsets, each having the same product of elements.

I think a contradiction should be useful here, and it should not be necessary to use this theorem by Erdos. So suppose $S$ can be partitioned into two subsets $A$ and $B$, each having the same products of the elements. Let $m$ be the smallest element in $S$. Then $\prod_{a\in A} a= \prod_{b\in B} b = \frac{1}2\prod_{j=1}^{p} (m+j-1).$ To get a contradiction, it might be possible to show that $B$ and $A$ cannot be disjoint, but I'm not sure how to do this. I know $S$ may or may not contain a multiple of $p$, but if one added another integer to $S$, then the new set must contain a multiple of $p$. Also, every nonmultiple of $p$ in $S$ has a unique inverse in $\mathbb{Z}_p^*,$ the multiplicative group of integers modulo $p$.

Any block of $p-1$ consecutive integers contains at most one multiple of $p$.

If it contains a (unique) multiple of $p$ then that term can only be in one of the two parts, so the two parts can't have the same product.

If it contains no multiples of $p$ then the block is a complete list of the non-zero residues $\pmod p$. We can then use Wilson's Theorem to deduce that the product of all the elements in the block is $-1\pmod p$. But, if your partition existed we'd have a solution to $n^2\equiv -1\pmod p$, which contradicts your assumption that $p\equiv 3 \pmod 4$.