Do the complex zeros of $\big(\zeta(s-1) -\zeta(s)\big) \pm \big(\zeta(1-s) - \zeta(2-s)\big)$ all (except 1 pair) reside on the line $\Re(s)=1$? Or stated differently: for $s \in \mathbb{C}$ and with $\chi(s)= \pi^{-s}\,2^{1-s}\,\cos\left(\frac{\pi\,s} {2}\right)\,\Gamma(s)$, do all, except a finite few, of the complex (real ones exist as well) zeros of:
$$\frac{\zeta(s-1)}{\zeta(s)}-\frac{\pm 1-\chi(s)}{\pm 1-\chi(s-1)}$$
reside on the line $\Re(s)=1$ ?
The finite few lying off the line are:


*

*$\pm = +$ the exceptional set: $(5.894... \pm 1.389...\,i)$ ,  $(2-
      5.894... \pm 1.389...\,i)$

*$\pm = -$ the exceptional set: $(3.006... \pm 2.438...\,i)$ ,  $(2-
      3.006... \pm 2.438...\,i)$


Could a proof for this be within reach or is it just as hard as the RH?
Thanks!
Added a graph of the + version on request.

 A: I do not think this is so difficult as the Riemann hypothesis, I will only
explain why this is so without giving complete proof.
First on the line $s=1+it$ the functions are
$$(\zeta(it)-\zeta(1+it))\pm (\zeta(-it)-\zeta(1-it)).$$
In other words $2\Re(\zeta(it)-\zeta(1-it))$ and $2i\Im(\zeta(it)+\zeta(1-it))$.
We have by the functional equation
$$\zeta(it)-\zeta(1-it)=(\chi(it)-1)\zeta(1-it).$$
For $t$ real and $t\to+\infty$ we have
$$\chi(it)-1=\Bigl(\frac{t}{2\pi}\Bigr)^{1/2}e^{i(-t\log\frac{t}{2\pi}+t+\frac{\pi}{4})}(1+O(t^{-1/2})).$$
The argument of $\zeta(1-it)$ is $O(\log t)$ [$O(\log\log\log t)$ under RH] and is zero at  points  $t_k\to+\infty$
(If taken $-\pi/2$ at $z=1$).
Therefore function $2\Re(\zeta(it)-\zeta(1-it))$ has approximately 
$\frac{T}{\pi}\log\frac{T}{2\pi}-\frac{T}{\pi}$ zeros in the interval $0<t<T$. 
That these are essentially all zeros of this functions need a little work, 
but I think it is possible to prove it by counting the number of all zeros
and comparing. 
The other function is treated analogously.
