Each of two summands decreases. It is equivalent to the fact that $(e^x-1)/x$ increases, and this follows from the convexity of exponent: if $f$ is convex, then $(f(x)-f(a))/(x-a)$ increases as a function of $x$ for fixed $a$, apply this to $f(x)=e^x$ and $a=0$.

I do not know wether this helps you or not, but you may do as follows.

Denote $f(x)=(e^x-1)/x$. Note that $f'(x)=\frac{fe^x(x-1+e^{-x})}{x(e^x-1)}>0$, so $f$ is a positive increasing function.

Lemma. The function $1/f=x/(e^x-1)$ is a (positive decreasing) convex function.

Proof. $$ (1/f)''=\frac{e^x(2+x-(2-x)e^x)}{(e^x-1)^3}, $$ we have to check that it is non-negative. If $x\geqslant 2$, this is clear. If $0<x<2$, this reduces to $$ e^x\leqslant \frac{2+x}{2-x}=1+x+x^2/2+x^3/2^2+x^4/2^3+\dots,\,\,(1) $$ that holds coefficient-wise: $n!\geqslant 2^{n-1}$ for $n\geqslant 1$. Finally, the case $x<0$ reduces to $x>0$, since $2+x-(2-x)e^x$ and $2-x-(2+x)e^{-x}$ always have opposite signs.

Corollary. $g:=1/f^2=x^2/(e^x-1)^2$ is convex.

Proof. $-(1/f^2)'=2(-1/f)'(1/f)$, both multiples are positive decreasing functions, thus their product also decreases.

Now we claim that $g(x-a)+x^2/(e^x+1)^2$ decreases for each $a\geqslant 0$, for $a=1$ we get your statement (and for other $a$ something equivalent to your remark). Since $g'$ increases, we have $g'(x-a)\leqslant g'(x)$, so it suffices to check this for $a=0$. We have $g(x)+x^2/(e^x+1)^2=2x^2(e^{2x}+1)/(e^{2x}-1)^2$. Denote $2x=y$, we need to check that $y^2(e^y+1)/(e^y-1)^2$ decreases. Taking logarithmic derivative, this is equivalent to $$\frac{2}x+\frac{e^x}{e^x+1}-\frac{2e^x}{e^x-1}\leqslant 0.$$ For $x=-y>0$ we have $$\frac2{y}-\frac2{e^y-1}\geqslant \frac2y-\frac2{y+y^2/2}=\frac1{1+y/2}>\frac1{1+e^y}$$ as desired. For $x>0$ we have $$ \frac{2}x+\frac{e^x}{e^x+1}-\frac{2e^x}{e^x-1}\leqslant \frac{2+x}x-\frac{2e^x}{e^x-1}= \frac{(2-x)e^x-(2+x)}{x(e^x-1)}\leqslant 0 $$ by (1).