Authors of the paper you linked actually define $f(z)$ differently. They have
$$
f(z)=\left(\frac{1}{1-z}-1\right)\zeta\left(\frac{1}{z-1}\right),
$$
so your $f(z)$ is their $f(-z^2)$ and every $\alpha$ in their formula corresponds to $\pm i\alpha$ from yours, so the sum on the left should actually be $\frac12$ of what is in the question, so
$$
\frac{1}{2}\left(\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|+ \sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|\right)=\sum_{-\pi/2<\arg(\rho)<\pi/2}\log\left|\frac{\rho}{1-\rho}\right|
$$
and everything works just fine, no contradiction.

P.S. And yeah, unfortunately the Riemann hypothesis is still hard.