Comparing two products of lists of integers?
(I understand the description of the problem so that the input numbers are bounded by a constant, so I will not track dependence on the bound.)
The problem is solvable in linear time and logarithmic space using sums of logarithms. In more detail, the algorithm is as follows:


Using binary counters, count the numbers of occurrences of each possible input number in both lists.


This takes time $O(n)$, and the counters use space $O(\log n)$, as each counter is bounded by $n$ in value.
Let $p_1,\dots,p_k$ be the primes below the $O(1)$ bound. By distributing each counter for a number $a$ to the prime factors of $a$ (with appropriate multiplicity), and subtracting the counts for one list from the other list, we obtain the following in time $O(\log n)$:


Compute integers $\beta_1,\dots,\beta_k$ with $O(\log n)$ bits such that the problem is equivalent to determining the sign of $\Lambda:=\sum_{i=1}^k\beta_i\log p_i$.

If $\beta_1=\dots=\beta_k=0$, answer that the products are equal.



Otherwise $\Lambda\ne0$. By Baker’s theorem, we can lower bound
$$|\Lambda|&gt;2^{-C\log n}$$
for a certain constant $C$. Thus, the following correctly computes the sign of $\Lambda$:


Output the sign of $\sum_{i=1}^k\beta_i\pi_i$, where $\pi_i$ is an approximation of $\log p_i$ to $m:=C\log n+k+1$ bits of accuracy.


Let $M(m)$ be the cost of multiplication of two $m$-bit integers. The current best bound is $M(m)=O(m\,\log m\,2^{O(\log^* m)})$, but here it would not make much difference even if we use the trivial $O(m^2)$ multiplication algorithm. We can compute $\log p_i$ to $m$ bits of accuracy in time $O(M(m)\log m)$ using AGM iteration (see e.g. here), and then evaluating $\sum_i\beta_i\pi_i$ takes time $O(M(m))$. Overall, step 4 takes time $O(M(m)\log m)\subseteq O(\log n\,\mathrm{poly}(\log\log n))$.
Thus, the running time of the algorithm is dominated by $O(n)$ of the first step.