While @reuns clearly showed the fallacy, the following is a simple explanation without getting bogged down in details that one can spin around in circles as we saw so often in this purported "proofs" here or on MO
The "proof" has the logical structure: RH is equivalent to the analyticity of $A(s), \Re s > \frac{1}{2}$ which is equivalent to the analyticity of $B(s), \Re s > \frac{1}{2}$.
We show that $A(s)-B(s)$ extends analyticaly to $\Re s > \frac{1}{2}$ hence we conclude RH (hence) that both $A(s), B(s)$ extend analyticaly to $\Re s > \frac{1}{2}$.
I think that this shows clearly the fallacy of the proof since for example $f(s)-f(s)$, where $f$ is any analytic functions on some domain, extends analytically to the whole plane...
Note first that the OP proof above actually shows that $I$ cannot be $\infty$ but it doesn't show that $I$ converges or it is not $-\infty$ since $I$ is an oscillating integral with $\log |\zeta|=-\infty$ infinitely many times on the critical line of course.
From general results about Hardy spaces and some changes of variable, it actually follows immediately that $I$ is finite but much more is known, namely that
$0 \le I \le 0.146$
(and much tighter bounds can be easily found by better numerics) as from the results of Balazard, Saias and Yor we know that:
$I=2\pi \sum_{\Re \rho >1/2}\log \frac{|\rho|}{|1-\rho|}=\pi \sum_{\Re \rho >1/2}\log \frac{|\rho|^2}{|1-\rho|^2}$
But now if $\rho=\sigma+it, \frac{|\rho|^2}{|1-\rho|^2}=\frac{\sigma^2+t^2}{(1-\sigma)^2+t^2}=1+\frac{\sigma^2-(1-\sigma)^2}{(1-\sigma)^2+t^2}$, so we have
$ 1 < \frac{|\rho|^2}{|1-\rho|^2} < 1+\frac{1}{(1-\sigma)^2+t^2}$ whenever $\sigma >1/2$ hence $0 < \log \frac{|\rho|^2}{|1-\rho|^2} < \frac{1}{(1-\sigma)^2+t^2}=\frac{1}{|1-\rho|^2}$
But now it is well known (see Broughan, Equivalents if The Riemann Hypothesis, vol 1, page 35) that: $\sum_{\rho} \frac{1}{|\rho|^2} \le .046191441$, where the sum is taken on all the non-trivial zeroes of zeta.
Since either $I=0$ if RH is true, or $0 <I < \pi \sum_{\Re \rho <1/2}\frac{1}{|1-\rho|^2}< \pi \times .046191441=0.145...$ if RH is false and the sum is non-void, we immediately see the claimed result; using that all the zeroes up to some huge $t$ are known to be on the critical line so not in the sum above, we can probably reduce the estimate of $I$ to a ridiculously low nonnegative number, showing the tightness of the equivalence.
Best Answer
This result involves a bunch of theorems in Hardy Space theory which can be found in the classic references by Duren or Garnett (chapter II especially)
I will quote the relevant theorems and then explain how they imply the result:
If $f \in H^p(\mathbb D), p>0$, then $f$ has non-tangential finite limit on the unit circle ae $d\theta$; if we call that $f^*(e^{i\theta})$, one has $\log |f^*| \in L^1(d\theta), |f^*| \in L^p(d\theta)$
There is a (unique up to constants of modulus one) decomposition $f=BSF$ where $F$ is outer (non-zero and determined by $\log |F^*|=\log |f^*|$), with $|F^*|=|f^*|$ ae, $B$ is a Blaschke product with the zeroes of $f$ inside the unit disc and $S$ is a singular inner function which is determined by a positive singular measure (wr $d\theta$) $\mu$; both $|B^*|=|S^*|=1$ a.e. $d\theta, |B(z)|, |S(z)| <1, |z|<1$ (unless either is trivial so identically $1$)
If $f$ is analytically continuable across some arc $I$ of the unit circle, then $B,S,F$ are so too (and of course vice versa) which concretely means that $B$ has no zeroes accumulating there and the measure $\mu$ has support disjoint from $I$
If the singular factor is non-trivial (ie measure $\mu \ne 0$, so $S$ is not $1$), then there is a point $e^{i\theta}$ for which $f(z) \to 0, z \to e^{i\theta}$ non-tangentially; this is very easy to see when $\mu$ is a point mass at some $e^{i\theta}$ so $S(z)=e^{-c\frac{e^{i\theta}+z}{e^{i\theta}-z}}, c>0$ hence $S$ decays exponentially at $e^{i\theta}$ while $F$ cannot increase too much since $F^*$ is in some $L^p$
Now for $f(z)=(s-1)\zeta(s), s=\frac{1}{1-z}$, one easily shows that it is in $H^p$ for all $p<1/2$ so it has a decomposition $f=BSF$ as above; we know that $(s-1)\zeta$ is analytically continuable across the finite part of $\Re s =1/2$ which corresponds with $f$ being so across $|z|=1, z \ne 1$; by Theorem 3 above, it follows that the singular measure $\mu$ which determines $S$ is concentrated at $1$ or trivial. If we were in the first case we would get $\lim_{r \to 1, r<1}f(r)=0$ (Theorem 4 above)
But now $x=\frac{1}{1-r}>0, x\to \infty, r \to 1$ and $f(r)=(x-1)\zeta(x)$ by defintion; but this would mean $\lim_{x \to \infty, x>0}(x-1)\zeta(x)=0$ and that is of course not true as the limit is infinity; hence $S=1$ and the measure $\mu$ is zero. Incidentally $B=1$ is equivalent of course with RH so $f$ is outer iff RH is true