Let $\Bbb{Q}^{\times} \xrightarrow{\Omega} \Bbb{Z}$ be the group hom given by $\Omega$ from the study of arithmetic functions (you can extend it uniquely (and homomorphically) to all of $\Bbb{Q}^{\times}$.)
It's kernel is generated by $\{\dfrac{p}{q} : p, q $ are prime numbers $\} = \ker \Omega$. Does the weaker-than twin prime conjecture:
$$ \ker \Omega \cap A = \ker \Omega \cap \{\dfrac{n}{n+2} : n \in \Bbb{Z}\} $$ is an infinite set hold?
Assume the intersection is finite, then $\ker \Omega \cap A' = \ker \Omega \cap \{\dfrac{n}{n-2} : n \in \Bbb{Z}\}$ is also finite. The set $B = \{ \dfrac{n^2}{n^2 - 4} : n \in \Bbb{Z}\} \subset AA'$ elemenwise. So that $\ker \Omega \cap B \subset \ker \Omega \cap AA'$. That's as far as I've gotten.
What happens elementwise with respect to $\cap$ is:
$$ A(B\cap C) \subset AB \cap AC $$
but not the other way around in general. So that:
$$ (\ker \Omega \cap A')(\ker \Omega \cap A) \subset (\ker \Omega)^2 \cap AA' \cap (\ker \Omega)A \cap (\ker \Omega)A' = \\ \ker \Omega \cap A\ker \Omega \cap A'\ker \Omega \cap AA' $$
So that in particular, the left side is a subset of $\ker \Omega \cap AA'$. So maybe this route is ruled out?
Note that $A^{-1} = A'$ by construction.
Other facts: $\Bbb{Q}^{\times} = \biguplus_{k \in \Bbb{Z}} p^k G$, where $G = \ker \Omega$, for any fixed prime $p \in \Bbb{P}$.
Take $A = \{\dfrac{n -1}{n+1} : n \in \Bbb{Z}, n \neq -1\}$, then similarly $G \cap A$ would be finite.
$$\dfrac{r_1\cdots r_i}{q_1\cdots q_i} = x \in G \cap A \iff \\ (n-1)q_1 \cdots q_i = (n+1) r_1 \cdots r_i, \\ q_j, r_j \in \Bbb{P} \\ \\ \iff \\ \Omega(\dfrac{\text{lcm}(n-1, n+1)}{n-1}) = \Omega(\dfrac{\text{lcm}(n-1, n+1)}{n+1}) \\ \iff \\ d := \gcd(n-1, n+1), \\ \Omega(\dfrac{n+1}{d}) = \Omega(\dfrac{n-1}{d}) $$
A paper by Goldston, Graham, Pintz, and Yıldırım gives, among other interesting results, that (Theorem 6 in this paper)
This gives a much stronger statement than you ask for; not only are there infinitely many rationals $\frac{x}{x+2}$ for which $$\Omega\left(\frac{x}{x+2}\right)= \Omega(x) - \Omega(x+2) = 0,$$ but there are infinitely many if you restrict the pair $(\Omega(x),\Omega(x+2))$ to be any $(B,B)$ with $B$ sufficiently large. This isn't quite enough to guarantee "twin semiprimes," but it's rather close.
Their results seem to be more general; in particular, it seems that they have also proven this where $2$ is replaced with any other positive integer, or where $\Omega$ is replaced with any sufficiently well-behaved number-theoretic function.