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

First recall the identity $$\int_0^{\infty} x^{({s-3})/2}\exp(-n^2\pi x)\,dx=\pi^{-(s-1)/2}\ \Gamma\left(\frac{s-1}{2}\right) n^ {-(s-1)}\,\mathrm{for}\, \Re(s)>2.$$ Multiplying 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 $\zeta$ relies on the fact that the function $\psi(x)=\sum_{n=1}^\infty{\exp(-n^2\pi x)}$ satisfies $$2\psi(x)+1={\frac{1}{x^{1/2}}}\left[2\psi\left(\frac{1}{x}\right)+1\right].$$ So maybe there is some similar property for $Q$ that lead to a functional equation for $F$.

About the zeros, since $\zeta$ has infinitely many zeros on the line $\Re(s)=1/2$, the obvious thing is that if your equation holds for $\Re(s)>0$, then $F$ must have infinitely many zeros on the lines $\Re(s)=3/2$ and $\Re(s)=1/2$

So, if $\rho$ is a zero of $F$ then $1-\rho$, $2-\rho$, and $1+\rho$ are also zeros of $F$.

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

First recall the identity $$\int_0^{\infty} x^{({s-3})/2}\exp(-n^2\pi x)\,dx=\pi^{-(s-1)/2}\ \Gamma\left(\frac{s-1}{2}\right) n^ {-(s-1)}\,\mathrm{for}\, \Re(s)>2.$$ Multiplying 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 $\zeta$ relies on the fact that the function $\psi(x)=\sum_{n=1}^\infty{\exp(-n^2\pi x)}$ satisfies $$2\psi(x)+1={\frac{1}{x^{1/2}}}\left[2\psi\left(\frac{1}{x}\right)+1\right].$$ So maybe there is some similar property for $Q$ that lead to a functional equation for $F$.

About the zeros, since $\zeta$ has infinitely many zeros on the line $\Re(s)=1/2$, the obvious thing is that if your equation holds for $\Re(s)>0$, then $F$ must have infinitely many zeros on the lines $\Re(s)=3/2$ and $\Re(s)=1/2$

So, if $\rho$ is a zero of $\zeta$ then $\rho$, $1-\rho$, $2-\rho$, and $1+\rho$ are also zeros of $F$.

This is not an answer but a quick remark about a possible direct proof of the 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 $re(s)>2$.

Multiplying 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 $F$.

About the zeros, since $\zeta$ has infinitely many zeros on the line $re(s)=1/2$, the obvious thing is that if your equation holds for $re(s)>0$ then $F$ must have infinitely many zeros on the lines $re(s)=3/2$ and $re(s)=1/2$

So, if $\rho$ is a zero of $F$ then $1-\rho$, $2-\rho$ and $1+\rho$ are also zeros of $F$.

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

First recall the identity $$\int_0^{\infty} x^{({s-3})/2}\exp(-n^2\pi x)\,dx=\pi^{-(s-1)/2}\ \Gamma\left(\frac{s-1}{2}\right) n^ {-(s-1)}\,\mathrm{for}\, \Re(s)>2.$$ Multiplying 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 $\zeta$ relies on the fact that the function $\psi(x)=\sum_{n=1}^\infty{\exp(-n^2\pi x)}$ satisfies $$2\psi(x)+1={\frac{1}{x^{1/2}}}\left[2\psi\left(\frac{1}{x}\right)+1\right].$$ So maybe there is some similar property for $Q$ that lead to a functional equation for $F$.

About the zeros, since $\zeta$ has infinitely many zeros on the line $\Re(s)=1/2$, the obvious thing is that if your equation holds for $\Re(s)>0$, then $F$ must have infinitely many zeros on the lines $\Re(s)=3/2$ and $\Re(s)=1/2$

So, if $\rho$ is a zero of $F$ then $1-\rho$, $2-\rho$, and $1+\rho$ are also zeros of $F$.

This is not an answer but a quick remark about a possible direct proof of the 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 $re(s)>2$.

Multiplying 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 $F$.

About the zeros, since $\zeta$ has infinitely many zeros on the line $re(s)=1/2$, the obvious thing is that if your equation holds for $re(s)>0$ then $F$ must have infinitely many zeroes on the lines $re(s)=3/2$ and $re(s)=1/2$

So, if $\rho$ is a zero of $F$ then $1-\rho$, $2-\rho$ and $1+\rho$ are also zeros of $F$.

This is not an answer but a quick remark about a possible direct proof of the 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 $re(s)>2$.

Multiplying 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 $F$.

About the zeros, since $\zeta$ has infinitely many zeros on the line $re(s)=1/2$, the obvious thing is that if your equation holds for $re(s)>0$ then $F$ must have infinitely many zeros on the lines $re(s)=3/2$ and $re(s)=1/2$

So, if $\rho$ is a zero of $F$ then $1-\rho$, $2-\rho$ and $1+\rho$ are also zeros of $F$.