Adding integers represented by their factorization is as hard as factoring? Reference request
Assume that we can solve the problem (lets call it FactSum) in complexity class $C$ and $C$ is closed under $\log$-iteration (aka $\log$-bounded recursion)(e.g. if we can compute $x*y$ where $*$ is a binary function, we can computed $x_1*\ldots*x_{\log{n}}$) and contains $\mathsf{P}$ (this last condition can be made weaker). We show that factoring is in also in $C$.

Note that each number can be written as a sum of $\log n$ powers of $2$. Each of them are easy to factor. 

Now given a number, write it as the sum of powers of it, then write each summand in the factoring representation, and then use the algorithm to sum them in the factoring representation. The result will be the factoring of the input number.

This shows that factoring is reducible to $\log$-iteration of your problem FactSum. Therefore factoring is in $\mathsf{P}^{\text{FactSum}}$ (and I think $\mathsf{P}$ can be replaced with  $\mathsf{NC^1}$ here).