Multiplicative version of 3-SUM
Your reduction from $3$SUM to $3$MUL works with a minor standard modification.
Suppose your original integers were in {$1,\ldots,M$}. After the transformation $x\rightarrow 2^x$ the new integers are in {$2,\ldots,2^M$}. We will reduce the range.

Consider any triple of integers $a,b,c$ in the new set $S'$.
The number of prime divisors of any nonzero $ab-c$ is $&lt;2M$. The number of such triples is $n^3$. Hence the number of primes $q$ which divide at least one of the $ab-c$ nonzero numbers is at most $2Mn^3$.

Let $P$ be the set of the first $2M\cdot n^4$ primes. The largest such prime is of size at most $O(Mn^4\log Mn)$. Pick a random prime $p\in P$. With high probability $p$ will not divide any of the nonzero $ab-c$, so we can represent each $a\in S'$ by its residue, mod $p$, and if $3$MUL finds some $ab=c$ in $S'$, with high probability it will be correct for the original $3$SUM instance. We have reduced the range of the numbers to {$0,\ldots, O(Mn^4\log Mn)$}.

(This is a standard size reduction. You might be able to do better by considering the fact that the $ab-c$ are always differences of two powers of $2$.)