This material appear related to a paper of Hurwitz: Ueber die Kettenbrüche, deren Teilnenner arithmetische Reihen bilden. (German) Zürich. Naturf. Ges. 41, 2. Teil. 34-64. Published: (1896). Also Lambert proved that $$\frac{e^u-1}{e^u+1}=[0,2/u,6/u,10/u,...]$$ this explain some of your numbers $2/r \in\bf Z$.

@Agno Your product is difficult to understand. For each zero $1/2+i\gamma$ (I assume that with $\gamma>0$) you add four zeros $\mu$, $1-\mu$, $\overline{\mu}$ and $\overline{1-\mu}$ to your product. But if RH is not true, and there are two zeros $\beta+i\gamma$ and $1-\beta+i\gamma$ (with $\beta > 1/2$ and $\gamma>0$) How many zeros have your product (associated with these two zeros) 4 or 8? Whatever the interpretation, I think that if RH is not true, your formula is not correct.

@Tom Leinster It is my attempt to give a sense the question of the OP. Presented as I have done it, you may insert this category under Peano axioms of Natural Numbers (as formulated for example in E. Mendelson, Introduction to Mathematical Logic, van Nostrand 1964). In this way you may think that it is part of a theory previous to Set Theory.

@RH The argument of $\Gamma(s)$ is well defined and harmonic on the plane with a cut along the negative real axis.

But I give the solution of the problem, assuming Riemann hypothesis. (I remind you that many believe this hypothesis to be true). The question appear to ask for a different treatment of zeros of zeta, according wheather they are on the line or not. As I said before with this interpretation (and negating RH) the question has no neat answer. My treatment is $\rho\mapsto 1/2+x(\rho-1/2)$ The question ask for $\rho\mapsto \Re\rho+x(\rho-\Re\rho)$. This is not an analytic function of ρ.

My formula do not need RH, but of course in case RH is false my gamma are complex and the scaling is not a movement on a vertical line for these zeros. If what you want is that the zero beta + i gamma move to beta + i x gamma, and RH is not true, I do not think there is a neat formula for the result.