I wanted to provide a clean derivation of a more precise formula of what you gave above. In particular, notice that that $\log\zeta(s)\approx P(s)$, so the $\zeta(s)$ and $\exp(-P(s)$ terms should cancel out in some way.

Let $\mathcal{R}=\{{n:\ \Omega(n)=2\}}$ and consider the logarithm of the product above. This equals $$\sum_{n\in \mathcal{R}} -\log(1-n^{-s})=\sum_{n\in \mathcal{R}}\sum_{k=1}^{\infty} \frac{n^{-ks}}{k}.$$ Setting $P_2(s)=\sum_{n\in \mathcal{R}} n^{-s}$, the left hand side above equals then equals $\sum_{k=1}^{\infty} P_2(ks)/k.$ Now, $$P(s)^2+P(2s)=\sum_{p}\sum_{q}(pq)^{-s}+\sum_{p}p^{-2s}$$ $$=2\sum_{p}\sum_{q\geq p} p^{-s}q^{-s},$$ and so $$\sum_{k=1}^{\infty} P_2(ks)/k=\frac{1}{2}\sum_{k=1}^\infty \frac{P(ks)^2+P(2ks)}{k}.$$ Now, $\log\zeta(s)=\sum_{k=1}^\infty P(ks)/k$, so we find that 


>$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\zeta(2s)\exp\left(\frac{1}{2}\sum_{k=1}^\infty P(ks)^2/k\right),$$ 

Now, in fact, $$\sum_{k=1}^\infty P(ks)^2/k=\sum_{q}\sum_{p}\sum_{k=1}^\infty\frac{q^{-ks}p^{-ks}}{k}$$ $$=\sum_{p}\sum_{q} -\log\left(1-(qp)^{-s}\right),$$ and hence

$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\zeta(2s)\sqrt{\prod_{p}\prod_{q}\left(1-(pq)^{-s}\right)^{-1}}$$ which I will rewrite as 

>$$\prod_{n:\Omega(n)=2}\frac{1}{1-n^{-s}}=\zeta(2s)\mathcal{P}$$

where $\mathcal{P}$ is precisely the double product appearing in your previous question http://mathoverflow.net/questions/190890/zeta-function-double-product.

To answer the main question "Is this close enough to be of use in any practical application?" my response is a pessimistic one, however this too vague of a question to give a concrete answer.