I think, [here](https://math.stackexchange.com/q/115230/19341), I found 
$$
P_x(s)=\sum_{p < x} \frac{1}{p^s} =\sum_{n=1}^{\infty}\frac{ \mu (n)}{n}
\sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}}
\left[
{\rm li}(t^{\frac zn-s})
\right]^{x}_2
 \tag{7}
$$

where $\rho$ are __all__ the zeros (trivial and non-trivial) of $\zeta$ function.  See the linked question for more detail, corrections are welcome.
Further we know, that
$$
    P(s)=\sum_{n> 0}\frac {\mu(n)}n{\log\zeta(ns)} .
$$

So my question is 

> If
$\lim_{x\to \infty} P_x(s)=P(s) $
then, by comparing coefficients, is
$$
\log\zeta(ns)=\lim_{x\to \infty}
\sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}}
\left[
{\rm li}(t^{\frac1n( z-ns)})
\right]^{x}_2
?
$$

What I got so far is:

* Could $  \displaystyle  \log \zeta(s) = s \int_0^\infty \frac{\pi(x)}{x(x^s-1)}\,dx $ be useful somehow?

* [Thanks to robjohn](https://math.stackexchange.com/a/285413/19341) it was possible to see that both coincide at least some special values:  
If $ns=1$ or $ns=\rho$, one addend in the sum diverges like $\lim_{x\to\infty} \log\left(\frac{\log(x)}{\log(2)}\right)=\infty$. So we get 
$$
\begin{eqnarray}
ns=1: & \log(\zeta(1)) &=& +\infty\\
ns=\rho: & \log(\zeta(\rho)) &=& -\infty 
\end{eqnarray}
$$

* I looked at the series expansion at $s=0$ for ${\rm li}(x^{\frac1n(z-ns)})={\rm Ei}((\frac zn-s)\ln(x))$ and $\log\zeta(ns)$. Assuming I'm not wrong, you'll get the following when you compare the linear terms
$$
\log(2\pi) \overset{?}{=} 
\lim_{x\to \infty}\sum_{z\in\{1,\rho\}}(-1)^{\delta_{1z}}
\left[ \frac{ t^{\frac{z}n}}{z}\right]_2^x ,
$$
which looks a little irritating, since the RHS has to be independent of $x$ an $n$. Does this show that it's wrong at all?