There are indeed many traps while evaluating $\pi(x)$ using the $\zeta$ zeros
(I will neglect here the smooth and quickly decreasing with $x$ contribution of the trivial zeros).
This answer detailed most of these traps but let's examine your specific points :
Now, whether I use $$\tag{3.1}R(x^\rho)=1+\sum _{n=1}^{\infty }
\frac{\big(\log(x^\rho)\big)^n}{nn! \zeta (n+1)}$$ or
$$\tag{3.2}R(x^\rho)=1+\sum _{n=1}^{\infty } \frac{\big(\rho
\log(x)\big)^n}{nn! \zeta (n+1)}$$ in $(1)$, in both cases I do not
get $\pi^*(x)$. (Series seems to diverge/oscillate.)
The known non trivial zeros $\rho$ verify the RH and may thus be written $\,\displaystyle\rho:=\frac 12+it$.
Now $(3.1)$ can't indeed provide the right answer because the numerical evaluation of $x^\rho$, before applying the logarithm, will give you for $x$ real $>1\quad\displaystyle x^\rho=\exp\left(\left(\frac 12+it\right)\log(x)\right)$.
The problem is that the evaluation of the logarithm will not return $\,\left(\frac 12+it\right)\log(x)\;$ but something like $\left(\frac 12+it\right)\log(x)+2\pi ki\;$ with $k\in\mathbb{Z}$ depending of $t$ (the evaluation of the logarithm imposing a phase in $(-\pi,\pi]\;$)
Using a different phase reference (branch) for the terms of the series isn't a fine idea but what about the Gram series as used in $(3.2)$?
This formula is indeed correct and was the one I used to produce this animation for $x\in(1,100)$. But this formula too has to be handled with care because we have to evaluate something behaving like (since $\,\zeta(n)\sim 1\,$ as $\,n\to +\infty$) :$$S(z):=\sum_{n=1}^{\infty } \frac{z^n}{nn!},\quad z:=\left(\frac 12+it\right)\log(x)$$
For the first non trivial zero and (say) $\,x=100\,$ we gave $|z|>65$, for the tenth $|z|>229$, for the hundredth $|z|>1089$.
The problem is that the terms of $S(z)$ and thus $R(x^\rho)\,$ will become very large (about $\dfrac {e^z}{z\sqrt{2 \pi z}}$ near $n=z$) before decreasing again to get under unity after about $\,\lceil e\,z\rceil\,$ terms. In practice for $\,x=100\,$ and the $100$-th non trivial zero you will need a working precision of $480$ digits!
Another idea proposed by reuns is to use the Möbius $\mu$ function formula (no high precision here) :
$$\tag{1}R(x)=\sum_{n=1}^{\lceil\log_2(x)\rceil} \frac{\mu(n)}n \operatorname{li}\bigl(x^{1/n}\bigr)$$
(since $\ \displaystyle\pi^{*}(x)=\sum_{n=1}^{\infty} \frac{\mu(n)}n \Pi^*\bigl(x^{1/n}\bigr)\;$ with $\Pi^*(z)=0\;$ for $\,z<2\;$ we need only the $\,\lceil\log_2(x)\rceil\;$ first terms for all the $R(x^\rho)$ expressions ... not just $R(x)$)
But from the logarithm in the $\operatorname{li}$ function we get exactly the same problem of "phase reduction" as previously. Fortunately a simply remedy exists from the definition of the exponential integral $\operatorname{Ei}$ by using $\,\operatorname{li}\left(e^x\right)=\operatorname{Ei}(x)$ and replacing $(1)$ with :
$$\tag{2}R(x)=\sum_{n=1}^{\lceil\log_2(x)\rceil} \frac{\mu(n)}n \operatorname{Ei}\left(\frac 1n\log(x)\right)$$
and
$$\tag{3}R\left(x^\rho\right)=\sum_{n=1}^{\lceil\log_2(x)\rceil} \frac{\mu(n)}n \operatorname{Ei}\left(\frac {\rho}n\log(x)\right)$$
The result for $\;\pi(x),\;x\in(2,100)$ (starting with $R(x)$ evaluated using the Gram series, subtracting the exact $\;\displaystyle\frac 1{\log(x)}-\frac 1{\pi}\arctan\left(\frac{\pi}{\log(x)}\right)\;$ contribution for the trivial zeros and two times the real part of the sum of the $\,200\,$ first non trivial zeros $\,R\left(x^\rho\right)\,$ terms using $(3)\;$) should be :
Concerning the use of the $\zeta$ zeros to evaluate the prime counting function $\pi(x)$ as in your expression $(1)$ this answer may he helpful.
Shortly the combinatorial Meissel-Lehmer method allowed to compute $\pi(x)$ up to $10^{23}$ while the analytic method allowed to go up to $10^{25}$ (in $2016$, I don't know the state of the art). Note that more efficient expressions than $(1)$ were used as you may find in the history by Büthe and references.
