Solution of an equation with Jacobi theta function

Let $\, g(x) := \theta_3(0,\mathrm{e}^{-ax^2 -b}).\,$ Your question about solutions to $\, x + x f(x) = 1 + f(1) \,$ is now about $\, x g(x) = g(1).\,$ Now $\,g(x)\,$ is a bell shaped curve with $\, g(x) > 0 \,$ and $\, g(-x) = g(x).\,$ If we can prove that $\,xg(x)\,$ is monotone increasing we are done. If it holds for $\,a=1, b=0\,$ then it holds in general.

In that special case, using Jacobi theta identities (mentioned by Mikhail Skopenkov in another answer), $\,xg(x) = g(\pi/x)\sqrt{\pi}$ and since $\,g(x)\,$ is monotone decreasing for $\,x>0,\,$ then $\,g(\pi/x)\,$ must be monotone increasing and we are done.

Have you tried Jacobi's identities for theta-functions? At least for $a=1, b=0$ the identity $$xg(x)=\sqrt{\pi}\,g(\pi/x)$$ implies that the function $xg(x):=xf(x)+x$ is monotone increasing.

Edit: the identity corrected, thanks to Somos.