$\newcommand{\mean}{\mathop{\mathrm{mean}}}$
Define $$ S(d) = \prod_{p|d\atop p>2}{p-1\over p-2}. $$ Bombieri and Davenport (1966) proved that $$ \mean\limits_{d\in{\mathbb N}} S(d) = \mean\limits_{d\in{\mathbb N}} \prod_{p|d\atop p>2}{p-1\over p-2} ~=~ \Pi_2^{-1} = 1 / 0.66016\ldots = 1.51478\ldots\,, \tag{1} $$ where $\Pi_2=0.66016\ldots$ is the twin prime constant.
It is not difficult to check that the average value of $S(d)$ remains unchanged ($=\Pi_2^{-1}$) if $d$ runs through an arithmetic progression $d=qn$ with $q=2^k$; in particular the average of $S(d)$ is $\Pi_2^{-1}$ when $d$ runs through all positive even integers.
Some experimentation with PARI/GP leads me to the following
Generalization of $(1)$ for an arithmetic progression $d=nq$, $n\in{\mathbb N}$: $$ \mean\limits_{d=nq\atop n\in{\mathbb N}} S(d) = \mean\limits_{d=nq\atop n\in{\mathbb N}} \prod_{p|d\atop p>2}{p-1\over p-2} ~=~ \Pi_2^{-1} \prod_{p|q\atop p>2}{p\over p-1}. \tag{2} $$
For example, if $q=5$, then average of $S(d)$ over the progression $d=5n$ is $$\mean\limits_{d=5n\atop n\in{\mathbb N}} S(d) = {5\over4}\Pi_2^{-1}.$$
Can anyone please point me to existing proofs of $(2)$? An idea/sketch of a proof would also be greatly appreciated!
Note: Let $q'$ be the largest odd divisor of $q$, then the "extra factor" in $(2)$ is
$$\prod_{p|q\atop p>2}{p\over p-1} = \prod_{p|q'}{p\over p-1} = {q'\over\varphi(q')}. $$ In particular, if $q$ itself is odd, then the "extra factor" is $q/\varphi(q)$, as was noted in comments.
