This is not an answer but a quick remark about a possible functional equation for $F$.

First recall the identity

$\int_0^{\infty} x^{({s-3})/2}exp(-n^2\pi x)dx\ =\pi^{-(s-1)/2}\ \Gamma\left(\frac{s-1}{2}\right)\ n^ {-(s-1)}$ for $re(s)>2$.

Multplying both sides by $1/(n+1)$ and summing for $n$ we get

$\int_0^{\infty} x^{({s-3})/2}Q(x)dx\ =\pi^{-(s-1)/2}\ \Gamma\left(\frac{s-1}{2}\right)\ F(s)$ 

Where $Q(x)=\sum_{n=1}^\infty\frac{exp(-n^2\pi x)}{n+1}$

Now the proof of the functional equation for $\zeta$ rely on the fact that the function
$\psi(x)=\sum_{n=1}^\infty\ {exp(-n^2\pi x)}$

Verify $2\psi(x)+1={\frac{1}{x^{1/2}}}(2\psi(\frac{1}{x})+1)$

So maybe there is some similar property for $Q$ that lead to a functional equation for $F$.

But in order for the equation $\zeta(s)=F(s)+F(s+1)$ to hold for $re(s)>0$ you can see that the fonctional equation for $F$ must relates values $F(s)$ to values $F(2-s)$ since it is already known that $\zeta$ has infinitely many zeroes on the line $re(s)=1/2$.

You then might expect that the functional equation for $F$ is in fact

$\pi^{-(s-1)/2}\ \Gamma\left(\frac{s-1}{2}\right)\ F(s)=\pi^{-(1-s)/2}\ \Gamma\left(\frac{1-s}{2}\right)\ F(2-s)$

So if your equation holds for $re(s)>0$ then $F$ must have infinitely many zeroes on the lines $re(s)=3/2$ and $re(s)=1/2$