I think, here, 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 $$ \log\zeta(ns)=\lim_{x\to \infty} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac1n( z-ns)}) \right]^{x}_2 ? $$

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

Thanks to robjohn it 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 $$ \begin{eqnarray} ns=1\to & \log(\zeta(1)) &=& +\infty \\ ; ns=\rho \to & \log(\zeta(\rho)) &=& -\infty \end{eqnarray} $$