Building on this question scaling the imaginary part of $\rho$s in infinite products, I like to conjecture that:

$$\displaystyle \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1- \frac{s}{1-\mu_n} \right)\left(1- \frac{s}{\overline{\mu_n}} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)$$

with $\mu_n = a + \Im(\rho_n)x i$ and $a,x \in \mathbb{R},x \ne 0, s \in \mathbb{C}$ and $\rho_n$ the n-th non-trivial zero of $\zeta(z)$,

has the following closed form:

$$\displaystyle H(s,a,x):= \frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} \frac{\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})}{\xi(\frac12 - \frac{a}{x}+ \frac{1}{x})}$$

where $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ is the Riemann xi-function.

If this formula is correct, the 'constructed' zeros $\mu_n$ can be stretched/condensed vertically via $x$ on the imaginary axis and shifted left/right on the real line via $a$. In all cases they would yield an entire function expressed by this closed form (think of it as a reversed application of the Weierstrass factorization theorem, i.e. starting with products of 'constructed' zeros).

Further factorization also seems possible with:

$$\frac{\xi(\frac12 - \frac{a}{x} + \frac{s}{x})}{\xi(\frac12 - \frac{a}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{\mu_n} \right) \left(1-\frac{s}{\overline{\mu_n}} \right)$$

and

$$\frac{\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})}{\xi(\frac12 - \frac{a}{x}+ \frac{1}{x})} = \prod_{n=1}^\infty \left(1- \frac{s}{1-\mu_n} \right) \left(1- \frac{s}{\overline{1-\mu_n}} \right)$$

When $a=\frac12$ and $x=1$, the formula correctly reduces to:

$$\frac{\xi(s)}{\xi(0)} = \prod_{\rho} \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)$$

from which the known Hadamard product for $\zeta(s)$ can been derived.

Unfortunately I do not have a proof for this formula, however rigorously checked it against many 'brute force' calculations using the first 2mln $\rho$s (all correct results, but accurate up to 5 decimals max). I manufactured the formula by replicating the symmetry of the closed form for $\mu_n = a + n x i$ (i.e. running through the integers rather than $\Im(\rho_n)$, see the linked question). Since until today, all non-trivial zeros appear to be lying on the critical line, I have used $\frac12$ as the "source" for all zeros for different $a$'s i.e.: $\frac12 - \frac{a}{x} + \frac{s}{x}$ just inserts $\frac12$ when $\Re(s)=a$. I guess I have thereby implicitly assumed the RH in constructing the formula.

My questions:

Is this a known closed form?

Does a proof of this closed form imply the RH, i.e. does it "force" the Hadamard product into a "straight jacket" that only allows it to be valid when all $a=\Re(\rho_n)=\frac12$ ?

UPDATE:

Assuming RH is true, I believe that I have found a nice proof for the equation in the OP. Since "the comments section is too small to contain it", I decided to put it as an answer to my own question to round it up.