Under What Conditions is $\prod_{n\in S}n^{1/n}\sim \prod_{n\in S}n^{1/|S|}$

Question

Suppose that $S\subset \mathbb{N}$. Under what conditions would we have that $$\prod_{\substack{n\in S \\ n\le x}}n^{1/n}\sim \prod_{\substack{n\in S \\ n\le x}}n^{1/|\{n\in S\;:\; n\le x\}|}\;?$$ Here $f\sim g$ means $\lim_{x\to\infty}f(x)/g(x)=1$. If it helps at all, my specific consideration is when $S$ is of density zero, i.e. $\lim_{x\to\infty}|\{n\in S\;:\; n\le x\}|/x=0$. Messing around with the relation we see that the above requires that $$|\{n\in S\;:\; n\le x\}|\sim \frac{\sum_{\substack{n\in S \\ n\le x}}\log n}{\sum_{\substack{n\in S \\ n\le x}}\frac{\log n}{n}}.$$ It may be that this is the simplest criterion.

As a note, the set of prime numbers satisfies this. This is because $$\sum_{p\le x}\frac{\log p}{p}\sim \log x\qquad \text{(Merten's First Theorem)}$$ $$\sum_{p\le x}\log p\sim x\qquad \text{(Chebyshev's Function)}$$ $$\pi(x)\sim\frac{\log x}{x}\qquad \text{(Prime Number Theorem)}$$ I am wondering if there is a simpler general condition for the above, since that would tell us that the first two above relations imply the third.