Let
$$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$
be the Riemann zeta function. The fact that we can analytically extend this to all of $\mathbb{C}$ and can find a zero free region to the left of the line $Re(s)=1$ shows that
$$\pi(x) := |\{p\leq x: p \mbox{ prime }\}| \sim \frac{x}{\log(x)}.$$
Moreover, improving the zero free region improves the error term on $\pi(x)$ with the Riemann hypothesis giving us the best possible error term.
However, there are elementary proofs for this fact about $\pi(x)$ which does not rely on using $\zeta(s)$. My question is, just using the knowledge that
$$\pi(x) = \frac{x}{\log(x)} + ET$$
for some error term I call $ET$, can you show that $\zeta(s)$ can be analytically continued with a zero free region to the left of the line $Re(s)=1$, where this region depends on $ET$.
For example, if you assume that $ET = x^{1-\epsilon}$ for some $\epsilon>0$, then can you show that $\zeta(s)$ can be analytically continued to the region $Re(s) > 1-\epsilon$ and that $\zeta(s)$ has no zeros in this region?
Any solution or reference would be greatly appreciated.
We have those Mellin transforms for $Re(s) > 1$ :
$$G(s) = \int_1^\infty (x-1) x^{-s-1}dx = \frac{1}{s-1}-\frac{1}{s}, \qquad F(s) = \int_1^\infty \frac{1-x}{\ln x} x^{-s-1}dx$$
We see that
$$F'(s) = G(s) \implies F(s) = \ln(s-1)- \ln(s) + C$$
Then use the Euler product, again for $Re(s) > 1$ :$$\zeta(s) = \prod_p \frac{1}{1-p^{-s}} \implies \ln \zeta(s) = \sum_p \sum_{k \ge 1} \frac{p^{-sk}}{k} = s \int_1^\infty J(x) x^{-s-1}dx$$
where $J(x) = \sum_{p^k \le x} \frac{1}{k}$ (see Abel summation formula, some sort of integration by parts).
Finally, it is easy to see that $J(x) = \pi(x) + \mathcal{O}(x^{1/2})$ so that $$\pi(x)- \frac{x}{\ln x} = \mathcal{O}(x^{\sigma}) \implies J(x)+\frac{1-x}{\ln x} = \mathcal{O}(x^{\sigma}) $$
$$ \implies \ln \zeta(s)+s\ln(s-1)-s \ln(s)+sC = s \int_1^\infty \left(J(x)+\frac{1-x}{\ln x}\right) x^{-s-1}dx$$
converges absolutely and hence is holomorphic for $Re(s) > \sigma \implies \zeta(s)$ has no zero on $Re(s) > \sigma$.