Let $S$ be a density zero set rational primes, in the concrete situation
$\#\{p<X,p\in S\}=\mathcal{O}(x/(\log x)^{3/2-\delta})$ for all $\delta>0$.


Then can $\prod\_{p\in S}(1-\frac{1}{p+1})$ be zero? If nonzero, do we have a lower bound? 



infinite product of (1-1/(p+1)) over a density 0 set of primes
Use $1-x \ge e^{-\frac{x}{1-x}}$ for $x \in (0,1)$ to establish 
$$\ln \prod\_{p \in S}\bigg(1-\frac{1}{p+1}\bigg) \ge -\sum\_{p \in S} \frac{1}{p}.$$


This is tight since $1-x=e^{-x + O(x^2)}$ in a neighborhood of $x=0$ and $\sum\_p \frac{1}{p^2}$ converges.


So one needs an upper bound on $\sum\_{p \in S} \frac{1}{p}$. We know $f(x)=\sum\_{p \in S, p <x} 1$, so this is a matter of summation by parts:


$$\sum\_{p \in S, p < x} \frac{1}{p} = \frac{f(x)}{x} + O\bigg(\sum\_{i<x} \frac{f(i)}{i^2}\bigg) = O\bigg(\frac{1}{(\log x)^{1.5 - \delta}}\bigg) + O\bigg(\sum\_{i < x} \frac{1}{i (\log i)^{1.5-\delta}}\bigg)$$


The main term is $O\bigg(\sum\_{i < x} \frac{1}{i (\log i)^{1.5-\delta}}\bigg)$ since $O\bigg(\frac{1}{(\log x)^{1.5 - \delta}}\bigg)$ goes to 0 as $ x \to \infty$. 


Luckily, the series $C\_{\delta}=\sum\_{i \ge 1} \frac{1}{i (\log i)^{1.5-\delta}}$ converges since $1.5>1$ - this can be shown by Cauchy condensation test, or by comparison to the following integral:
$$\int\_{e}^{\infty} \frac{dx}{x(\log x)^{1.5-\delta}} = \int\_{1}^{\infty}\frac{dt}{t^{1.5-\delta}} = \bigg(\frac{1}{2}-\delta\bigg)^{-1}.$$


This proves your product doesn't vanish and gives the lower bound $\exp(-O(C\_{\delta}))$ for any $\delta >0$.


