Conjecture.
For any integer $x \geq 1$, we have: $$ A(x) = \prod_{p,q \in \Bbb{P} \\ p \lt q} \left(1 - \frac{x^2 - 1}{pq}\right) = 0 $$
if and only if $x$ is a twin prime average.
How can we prove that $A(x)$ either diverges or does not converge to $0$ if $x$ is not a twin prime average?
The $p \lt q$ is not necessary, as $x$ is assumed always an integer; see comments.
I think, but I'm not certain that $A(x)$ can be written:
$$ A(x) = 1 + (x^2 - 1)\lim_{N \to \infty} \ \left(\sum_{n = 4, \ \\ 2\ \mid \ \Omega(n)}^{N} \dfrac{(-1)^{\Omega(n) / 2}}{n}\right) $$
That is my attempt so far. I am not sure if it's correct or not.
The core problem with this infinite product is that $$ \sum_{\substack{p<q\\ \text{prime}}}\frac{1}{pq} =\frac{1}{2}\left[ \left(\sum_{p\,\text{prime}}\frac{1}{p}\right)^2 -\sum_{p\,\text{prime}}\frac{1}{p^2} \right] = \infty \tag{1} $$ as the sum $\sum_{p\,\text{prime}}1/p=\infty$ while $\sum_{p\,\text{prime}}1/p^2$ converges.
If you plug in $\epsilon=1-x^2$ and take the logarithm to convert the product into a sum, $$ \ln A(x) = \sum_{\substack{p<q\\ \text{prime}}} \ln\left(1+\frac{\epsilon}{pq}\right) = \sum_{\substack{p<q\\ \text{prime}}} \left(\frac{\epsilon}{pq} + \frac{O(\epsilon^2)}{p^2q^2}\right) \tag{2} $$ where the $O(\epsilon^2)$ term is bounded. So the sum diverges to $\pm\infty$ with sign depending on the sign of $\epsilon$.
Thus, if $|x|>1$, then $\epsilon<0$ and the sum becomes $-\infty$ making $A(x)=0$; if $|x|<1$, then $\epsilon>0$ and the sum becomes $+\infty$ making $A(x)=\infty$.
However, if you increase the power in the denominator $$ F_s(t) = \prod_{\substack{p<q\\ \text{prime}}} \left( 1-\frac{t}{(pq)^s} \right) $$ it will converge when $s>1$. You can then define $A_s(x)=F_s(-(x^2-1)^s)$ which should have the desired convergence and integer valued roots corresponding to twin-primes.