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I found that the following infinite product with $\mu = a +n b i$ and a,b real, $s \in \mathbb{C}$:

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

can be expressed in a closed form (with poles at $a,b = 0$ or $a=1$ and when $a=s$):

$$\dfrac{\left( {a}^{2}-a \right)} {\left( {a} ^{2}-a+s-{s}^{2} \right)} \dfrac{\Gamma \left( {\frac {-ia}{b}} \right) \Gamma \left( {\frac {-i \left( a-1 \right) }{b}} \right)}{\Gamma \left( {\frac {-i \left( a-s \right) }{b}} \right) \Gamma \left( {\frac {-i \left( a+s-1 \right) }{b}} \right)}$$

When $a=\frac12$ this could be further reduced to (poles at $s=\frac12$ and $b=0$):

$$\dfrac{1}{(2s-1)} \dfrac{\sinh \left( {\frac { \left( 2s-1 \right) \pi }{2b}} \right)} { \sinh \left({\frac {\pi }{2b}} \right)}$$

Encouraged by this result, my wish was to use it to find new hints about the Hadamard product:

$$\displaystyle \prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right) = \dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$$

but it is quite obviously an impossible task to transform the linear element $nb$ into to very random imaginary parts of the $\rho$s. However, it still triggered a follow up question:

With $\rho = \sigma + ti$ and $t,x$ real, the following product:

$$Had(s,x):=\displaystyle \prod_\rho \left(1- \frac{s}{\sigma + xti} \right) \left(1- \frac{s}{1-(\sigma + xti)} \right)$$

allows for "scaling" of the imaginary parts of the $\rho$s.