From this question, it is easy to derive that a zero of $\xi(a+s)\pm \xi(a+1-s)$ should occur when:
$$\displaystyle{\frac {\zeta \left( s+a \right) }{\zeta \left( s-a \right) }}=\pm{ \frac {{\pi }^{-\frac12+a}\,{2}^{a+1-s}\cos \left( \frac12\,\pi \, \left( s-a \right) \right) \,\Gamma \left( \frac12\,a+\frac32-\frac12\,s \right) \,\Gamma \left( s-a+1 \right) }{ \left( a+s-1 \right) \,\Gamma \left( \frac12\,a+\frac1 2\,s+1 \right) }}$$
For $a=1$, I found that this can be quite drastically reduced to:
$${\frac {\zeta \left( s+1 \right) }{\zeta \left( s-1 \right) }}=\pm\, 2\,\pi\frac{2-s}{s\,(s+1)}$$
Given the simplicity of this equation, could there be a way to proof that all complex solutions to this particular equation do reside on the critical line $\Re(s)=\frac12$?
Addition
I believe the question could be simplified even further by using $\xi^*(s) =\pi^{-\frac{s}{2}}\, \Gamma\left(\frac{s}{2}\right)\, \zeta(s)$ instead of $\xi(s)$, so that a complex zero in the strip of $\xi^*(a+s)\pm \xi^*(a+1-s)$ is induced when:
$${\frac {\zeta \left( s+1 \right) }{\zeta \left( s-1 \right) }}=\pm\frac{2\,\pi}{s-1}$$
Since $0<\Re(s)<1$, this can be rewritten as:
$${\frac {s-1 }{\zeta \left( s-1 \right) }}\cdot \prod_{primes}\frac{1}{1-\frac{1}{p^{s+1}}}=\pm\,2\,\pi$$
It is really nice to see how each incremental prime added to a finite Euler product, brings the $\Re(s)$ of a zero closer and closer to $\frac12$.