Richard Mathar $[1]\& [2]$ shows that \begin{align} &\zeta_{2}(s)\equiv\prod_{\Omega(n)=2}^{}\left(\dfrac{1}{1 - n^{s}}\right)^{-1}=\exp \left(\sum _{k=1}^n \frac{P(k s)^2+P(2 k s)}{2 k}\right)\\ \end{align} where $P(s)$ denotes the prime zeta function on $s,$ and $\Omega(n)=2,\ n\in\mathbb{N}$ the set of semiprimes.
Although of interest in itself, this identity doesn't give much insight into the expansion of the Euler product in question. Taking \begin{align} &\zeta(s)\prod_{\Omega(n)=2}^{}\left(\dfrac{1}{1 - n^{-s}}\right)^{-1}\\ \end{align} and sieving \begin{align} &\left(1-\frac{1}{4^s}\right)\zeta(s) = 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{6^s}+\frac{1}{7^s}+\frac{1}{9^s}+ \ldots\\ \\ &\dots\text{etc.} \end{align} leads to the first few terms $\left(\text{excluding the obvious }1+\sum_{p\in\mathcal{P}}\dfrac{1}{p^s}\right)$ \begin{align} &-\frac{1}{12^s}-\frac{1}{18^s}-\frac{1}{20^s}-\frac{1}{28^s}-\frac{2}{30^s}-\frac{1}{36^s}-\frac{2}{42^s}-\frac{1}{44^s}+ \ldots\\ \end{align} where the numerator is clearly characteristic of the prime factorization of the denominator. Taking only the primorial denominators from $\#{3}$ upwards and ignoring the $s$ exponent: \begin{align} &-\frac{2}{30}-\frac{2}{210}+\frac{6}{2310}+\frac{16}{30030}-\frac{20}{510510}-\frac{132}{9699690}+\frac{28}{223092870}+ \ldots\\ \end{align} which are clearly the coefficients of the Taylor expansion of $\exp(x-x^2/2)$
CoefficientList[Series[Exp[x - x^2/2], {x, 0, 10}], x] Range[0, 10]!
giving \begin{align} &\prod_{\Omega(n)=2}^{}\dfrac{1}{1 - n^{-s}}\approx\zeta (s)\exp \left(P(s)^2/2-P(s)\right)\\ \end{align} which agrees with the identity given at the beginning of the post.
Why do the sieved values of the primorial terms add up to precisely the coefficients of the Taylor expansion? I am sure there is a simple combinatorial explanation for this, but for now, it eludes me.
$[1]$Hardy-Littlewood Constants Embedded into Infinite Products ... Richard J. Mathar
$[2]$Series of Reciprocal Powers of k-almost Primes Richard J. Mathar
