Consider the prime counting function: $$\mathrm{Zi}(x)=\frac{1}{e}\sum_{k=1}^\infty\frac{(\ln x)^k}{kk!\phi(k)}$$
where $$\phi(k)=\sum_{n=1}^\infty e^{-{n^k}}$$
What happens when you follow this prescription for $\mathrm{Zi}(x)?$ Do you exactly obtain $\pi(x)?$ Is there some other set of "nontrivial zeros" that would give exactly $\pi(x)?$
$$ \mathrm{Zi}_n(x)=\mathrm{Zi}(x)-\sum_{m=1}^\infty \mathrm{Zi}(x^{-2m})+ \sum_{k=1}^n T_k(x) $$
where $$ T_k(x)= - \mathrm{Zi}(x^{\rho_{k}})-\mathrm{Zi}(x^{\rho_{-k}}) $$
My guess is that in the limit you'd obtain a function very close to the prime counting function $\pi(x)$ but it wouldn't be exact.
Here is a plot of $\pi(x)$ and $\mathrm{Zi}(x):$
