I'm trying to give a proof. This proof is of course not elegant, but it works!
For $x\le 0$, let $g(x) = e^\frac{x^2}{2}-e^x+x$, then $g'(x) = xe^\frac{x^2}{2}-e^x+1$, $g''(x) = (x^2+1)e^\frac{x^2}{2}-e^x\ge (x^2+1)e^\frac{x^2}{2}-1\ge 0$. So $g'(x)$ increases on $(-\infty,0]$, hence $g'(x)\le g'(0)=0$ for all $x\le 0$; $g(x)$ decreases on $(-\infty,0]$, so $g(x)\ge g(0)=0$ for all $x\le 0$. This shows that $e^x\le x+e^\frac{x^2}{2}$ for $x\le 0$.
Now suppose that $x\ge 0$. Note that $e^x\ge 1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^n}{n!}$ for all odd $n$ and $x\in\mathbb{R}$, so we have
$$e^{\frac{9}{16}x^2-x} + xe^{-x}-1\ge 1+\left(\frac{9}{16}x^2-x\right)+\frac{1}{2}\left(\frac{9}{16}x^2-x\right)^2+\frac{1}{6}\left(\frac{9}{16}x^2-x\right)^3+x\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{120}\right)-1,$$
which simplifies to
$$e^{\frac{9}{16}x^2-x} + xe^{-x}-1\ge \frac{x^2}{122880}(2621x^4 - 14320x^3 + 33520x^2 - 28160x + 7680).$$
It remains to show that $2621x^4 - 14320x^3 + 33520x^2 - 28160x + 7680\ge 0$. Well, there are several ways to prove this, for example by noting that
$$2621x^4 - 14320x^3 + 33520x^2 - 28160x + 7680 = \frac{1}{10484}(5242x^2-14320x+7049)^2 + \frac{3}{15826376437}(6038297x-3889420)^2 + \frac{1763413283}{24153188}.$$
