I'm not sure this really constitutes an answer, but I think that it may not be fruitful to study the type of generalized question you discussed at the start of your post; you should be able to prove rather easily that for any real number $r>1$ you can find a sequence of primes $p_i$ with exponents $e_i=p_i$ such that the infinite product $\prod_{i=1}^\infty (p_i)^{1/e_i}$ converges to exactly $r$. Just pick the primes in a greedy fashion. Or maybe you use every prime, and choose the exponents in a greedy fashion. Or mix and match, and end up with many ways to represent your real in this fashion.

Given that there is so much choice and so little effect, it's my opinion that you're going to have to rely on the specific structure of your product to go anywhere fast, and perhaps this isn't so unreasonable of an opinion given the nature of almost all results we have about transcendence (namely that they exploit the very special definitions, structures, relations, etc., that their subject numbers enjoy, like $e$ and $\pi$). But hey, I'm no expert.

Edit: New account, can't make comments, so this space is for Wojowu's request (and perhaps others in the future).

You're right in questioning me on that point; this is not a proof that I have done myself, but it seems fairly straight forward given that the product $\prod_{i=1}^\infty (p_i)^{1/p_i}$ diverges if you take it over all primes. You can't fail to make it past $r$ since you can always just add more primes in sequence (there is always a "partial tail" that is as large as you want), but by selectively (greedily) deleting certain primes we can lower the partial products below $r$. So you should converge exactly to $r$. If you go about formalizing that and run into problems, let me know.