Partial product of Euler factors Let $\mathbb P$ denote the set of prime numbers and for a subset $T\subset \mathbb P$ let
$$
\zeta_T(s)=\prod_{p\in T}\frac1{1-p^{-s}},
$$
where $\mathrm{Re}(s)>1$.
Is there any $T$ such that $T$ and $T^c={\mathbb P}\smallsetminus T$ are both infinite and $\zeta_T$ has a meromorphic continuation to $\mathbb C$?
 A: (Not an answer but a long comment.)  I think it is a safe bet that there are no such functions, but this possibility will not be easy to rule out because for a 
Dirichlet series being meromorphic is not a very handy condition.
I would like to point out some obvious but probably interesting facts about this hypothetical decomposition. We have 
$$\zeta(s)=\zeta_{T}(s)\zeta_{T^c}(s),$$
and $\zeta(s)$ has a pole at $s=1$. For both functions 
$$\zeta_{T}(s)>1,\,\zeta_{T^c}(s)>1,\,s>1$$
hence one of them has a pole at $s=1$ (the "big" one) and another does not (the "small" one). We may assume that $\zeta_{T}(s)$ is the latter.
The Dirichlet series 
$$\zeta_{T}(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$
has nonnegative coefficients ($a_n\ge 0$), hence if it does not have any poles on the line $s>0$, then both the series and the product converge absolutely on the halfplane $\Re s>0$. It is not difficult to see that in such a case this function cannot have an analytic continuation beyond $\Re s=0$; the reason for this are the poles of the factors $\frac{1}{1-p^{-s}}$ on this line. 
For a proof of this fact it is convenient to use the logarithmic derivative
$$-\frac{\zeta'_{T}(s)}{\zeta_{T}(s)}=\sum_{p\in T}\frac{\log p}{p^s-1},\,\Re s>0.$$ 
If $\zeta_{T}(s)$ were meromorphic at $s=0$ then we would have
$$-\frac{\zeta'_{T}(s)}{\zeta_{T}(s)}\sim \frac{n}{s},\,s\to+0$$
but RHS grows faster then this (for infinite $T$).
It follows that the set $T$ can't be too small, there is an asymptotic like
$$\#\{p\in T: p\le x\}\sim n\int_2^x\frac{dt^\sigma}{\log t}$$
($0<\sigma<1$), which can be proved more or less the same way as the prime number theorem.
