Non trivial zeros of the Zeta function The Zeta-function can be written as the following infinite Hadamard product of its non-trivial zeroes: 
$\zeta(s) = \pi^{\frac{s}{2}} \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$
this also implies that:
$\zeta(1-s) = \pi^{\frac{(1-s)}{2}} \dfrac{\prod_\rho \left(1- \frac{(1-s)}{\rho} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$
Take the reflection formula: 
$\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$
and substitute the Hadamard products for $\zeta(s)$ and  $\zeta(1-s)$. The result is that:
$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$
that can be rewritten as:
$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$
This equation has zeros for $\rho = s$ as long as $2\rho-1 \ne 0$.
The ρ's could obviously lie anywhere in the already proven strip between $0<\Re(\rho)<1$ and I just take them as 'givens' wherever they might be located. 
I started to experiment with solving s for different numbers of terms as follows:
$x=\frac13$ (e.g.)
$\prod_{n=1}^y \left(\frac{(x + ni) -s}{(x + ni) + s -1} \right) = 1$
Whatever value I pick for x between 0 and 1, the solution is always a complex number (ignoring $s=\frac12$, that is always a solution). However the exception occurs when $x = \frac12$ that always seems to only produce real numbers as solution(s). 
Does anybody see why that is (or must be) the case?
 A: Using the notation $s=u+1/2$ your conjecture can be reformulated and generalized as follows. 
Proposition. Let $v_1,v_2,\dots,v_N$ be arbitrary positive numbers, then all solutions of the equation
$$ \prod_{n=1}^N \frac{v_ni-u}{v_ni+u} = 1 $$
are real.
Proof. The degree of the polynomial $\prod_{n=1}^N(v_ni-u)-\prod_{n=1}^N(v_ni+u)$ is $N$ or $N-1$ depending on whether $N$ is odd or even (the polynomial is always odd). Therefore it suffices to show that there are the same number of real solutions to the displayed equation. As $u$ grows from $-\infty$ to $\infty$, each fraction under the product traverses the unit circle continuously in the positive direction, starting from and arriving back to $-1$. Using ideas similar to how one proves that the fundamental group of the unit circle is $\mathbb{Z}$, we see that the product traverses the unit circle $N$ times in the positive direction, starting from and arriving back to $(-1)^N$. In particular, the product passes $1$ exactly $N$ or $N-1$ times depending on whether $N$ is odd or even. QED
A: I would like to go back to my original question one more time.
$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$
that can be rewritten as:
$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$
The insight I gained from the various comments (thanks GH), is that this equation can not be used to 'solve' $s$ by taking the $\rho$'s as given inputs. The equation is just valid for all $s \in \mathbb{C}$ and it is incorrect to reverse the argument by for instance assuming that all numbers can be produced as solutions from the $\rho$'s.
So, another approach is to assume that the $s$ is a given input (arriving via the $\zeta(s)$) and that the $\rho$'s are the solutions for any $s$ chosen from $\mathbb{C}$. To avoid a too strong link with the non-trivial zeros (that encode specific info about the prime numbers only and therefore are expected to be a very specific subset of all solutions for this equation), I use the variable $x$ instead of the $\rho$.
This gives the following equation:
$\prod_{n=1}^N \frac{x_n-s}{x_n + s -1} = 1$
Experimenting with various input values for $s$, I now dare to conjecture the following:
1) $s \in \mathbb{R}$ always produces $x_n = \frac12 \pm yi$; $y \in \mathbb{R}$. EDIT : Additional note:


*

*for $s \in \mathbb{R}, s \not \in
   \mathbb{Z}$, all solutions are
perfectly symmetrical ie.: $x_n =
   \frac12 \pm yi$.

*for $s \in \mathbb{Z}$, (most)
solutions are of the form $x_n =
   \frac12 + yi$, otherwise symmetrical.


2) $s \in \mathbb{C}$ and $s= a \pm yi$ and $a \ne \frac12$ always produces $x_n = \frac12 \pm (w + yi)$; $w \in \mathbb{R}$
3) $s \in \mathbb{C}$ and $s= \frac12 \pm yi$ always produces $x_n = \frac 12 \pm w$; $x_n, w \in \mathbb{R}$
The last outcome is easy to prove by taking GH's proof from the post above and assuming $s=\frac12 + v_n i$ (or $-v_n i$) and $x = \frac12 + u$ (or $-u$).
Proving the first and middle outcomes is much more difficult, however I believe a potential approach for the first result could be to reverse GH's proof and maybe assume a bijective relationship like this: $x_n = \frac12 + y i \rightarrow s \in \mathbb{R}$, and therefore: $s \in \mathbb{R} \rightarrow x_n = \frac12 + y i$.
These maybe just some lose observations about the types of solutions for $x_n$ (of which the $\rho$'s are assumed to be a subset), however it does yield a direct conflict with the Riemann hypothesis. Assuming it is proven that $s= \frac12 \pm yi$ always produces a real result for $x_n$ (and therefore $\rho_n$), the $\rho$ could then never become equal to $s$ anymore (and turn the left or right term in the original equation and thereby $\zeta(s)$ into zero). But maybe the way around this dilemma is to assume that the zeros only occur in a limit situation of e.g. $\lim_{a \to \frac12} x_n = a \pm yi$) 
But I guess I'm doing something wrong here and would appreciate any guidance on where the logic derails.   
A: Just wanted to share my latest train of thoughts and leave you with the open question if the logic makes sense or not.
The following equation can be derived from combining the Hadamard products of the non-trivial zeros ($\rho$) from the Zeta function and its reflection formula:
$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{\rho} \right)$
that can be rewritten as:
$\prod_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$
This equation must be valid for all $s$ (except 1) and to learn more about the $\rho$'s, I tried to solve the following more generic (and less infinite) equation: 
$\prod_{n=1}^N \left(\frac{x_n-s}{x_n + s -1}\right) = 1$ 
It can be proven (see above) that when $s= \frac12 \pm yi$ the equation always produces $x_n = \frac 12 \pm w$; $x_n, w \in \mathbb{R}$. The outcome is therefore always real and this is in direct conflict with the empirical evidence that at least the first billions of non-trivial zeros lie on the (complex) critical line. The situation that $s$ = $\rho_n$ (i.e. a non-trivial zero, subset of $x_n$) can therefore never be achieved and this makes me suspicious something is wrong here.
Therefore the derived equation must be incorrect and the only decent way out of it I see is to assume the following subtle change:
$\zeta(1-s) = \pi^{\frac{(1-s)}{2}} \dfrac{\prod_\rho \left(1- \frac{(1-s)}{(1-\rho)} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$
Assumption (A):
$\rho = a + yi$ and $(1-\rho) = (1-a) - yi$ or $\rho = a - yi$ and $(1-\rho) = (1-a) + yi$.
$\prod_\rho \left(1- \frac{s}{\rho} \right) = \prod_\rho \left(1- \frac{(1-s)}{(1-\rho)} \right)$ 
that can be dramatically simplified into (note the $s$ dropping out):
$\prod_\rho \left(\frac{\rho -1}{\rho} \right) = 1$
and more generically to enable experimenting:
$\prod_{n=1}^N \left(\frac{x_n-1}{x_n}\right) = 1$
Solving this equation for all $x_n$ being equal, the outcome is always $\frac12 \pm y i$, so that's a much more promising outcome for inducing zeros of the Riemann hypothesis at $s=\rho_n$.
But we know that each $\rho_n$ is different and could lie anywhere in the critical strip $0<\Re(\rho)<1$. Obviously by simply assuming that all $\rho$'s (as a subset of $x_n$) are lying on the critical line, it is simple to proof that the equation nicely holds, since each (absolute) term in the product will be equal to 1:
$|\prod_{n=1}^N \left(\frac{\frac12 + ni -1}{\frac12 + ni}\right)| = 1$

N.B: Just briefly like to share a nice
  byproduct I observed when playing with this
  equation: 
$k_n, a, b, c \in \mathbb{Z},
> |\prod_{n=1}^N \left(\frac{\frac12 +
> k_n i -1}{\frac12 + k_n i}\right)| =
> |\frac{a}{c} + \frac{b}{c}i| = 1
> \rightarrow a^2 + b^2 = c^2$ 
Only Pythagorean triples will be produced
  for random values of $N$ and $k_n$.

So what will happen when some $\rho$'s are lying off the critical line? This would make at least two terms not being 1 anymore (one that causes it and one complementary to make the total product equal to 1 again). Example:
$\left(\frac{\frac13 + 6i -1}{\frac13 + 6i}\right)\left(\frac{y -1}{y}\right)=1$ gives $y=\frac23 -6i$.
But here's the trick: assumption (A) above now prohibits the $\rho$'s to switch signs in the product (only $(1-\rho)$ does that). This therefore implies that only $\rho_n = \frac12 + yi$ or $\rho_n = \frac12 - yi$ can be valid solutions for the equation. 
And now I only have to proof assumption (A) is true... 
A: My last thought on this one (I promise).
To be more precise, I indexed the non-trivial zeros in the formula in the opening post but now changed the $\rho$ in the second product into $1-\rho$ (that is in line with Riemann's observation that when a $\rho$ is a non-trivial zero, also $1-\rho$ must be one):
$\displaystyle \prod_{\rho_1}^{\rho_\infty} |\left(1- \dfrac{s}{\rho_n} \right)|=\prod_{\rho_1}^{\rho_\infty} | \left(1- \dfrac{1-s}{1-\rho_n} \right)|$
This means that each term in the product with the same $\rho_n$ can be equated as follows:
$|\left(\dfrac{{\rho_n} - s}{\rho_n} \right)| = |\left(\dfrac{s-\rho_n}{1-\rho_n} \right)|$
EDIT: This step is cleary not allowed and implicitly already assumes $\Re(\rho_n) =\frac12$. The terms in both products can be different. Dividing out all terms with equal $\rho_n$ is allowed (making the infinite product $1$) as I did in my previous post, but this doesn't yield any additional info about the individual $\rho_n$. So, back to the drawing board.
This equation is valid for all $s$ when $\Re(\rho_n) =\frac12$, but it is also valid for all $s=\rho_n$. And that could be for any complex number $\rho_n$ in the critical strip. However, to better see what happens when $s$ approaches $\rho_n$, the following equation gives an indeterminate form of type $0/0$: 
$|\left(\dfrac{{\rho_n} - s}{s-\rho_n} \right)| = |\left(\dfrac{\rho_n}{1-\rho_n} \right)|$ or $|\left(\dfrac{{s-\rho_n}}{\rho_n -s} \right)|= |\left(\dfrac{1-\rho_n}{\rho_n} \right)|$
but by applying L'Hôpital's rule we find, 
$\displaystyle \lim_{s \to \rho_n} |\left(\dfrac{{\rho_n} - s}{s-\rho_n} \right)| =1$
which implies that when $s$ approaches $\rho_n$ infinitely close, the following equation must be true:
$|\left(\dfrac{\rho_n}{1-\rho_n} \right)| = |\left(\dfrac{1-\rho_n}{\rho_n} \right)| =1$
And this equation only has solutions when $\Re(\rho_n) =\frac12$. It also implies (if the logic is correct) that each term in the following infinite products:  
$\displaystyle \prod_{\rho_n} |\left(\frac{1- \rho_n}{\rho_n} \right)| = 1$
or
$\displaystyle \prod_{\rho_n} |\left(\frac{\rho_n}{1-\rho_n} \right)| = 1$
is equal to $1$ and therefore all $\rho$'s contribute independently from each other to the overall product (and are therefore simple?).
