The inequality $e^y \ge 1+y$, holds for all $y \in \mathbb{R}$. Assume the function $g$ is monotonously increasing on $\mathbb{R}$. Then, based on the inequality above, we get
$$f(x + y) = e^{x+y}g(x + y) \ge e^x(1 + y)g(x) = (1 + y)f(x),$$
for all $x \in \mathbb{R}$ and all $y \ge 0$.

Conversely, assume $f(x + y) \ge (1 + y)f(x)$, for all $x \in \mathbb{R}$ and all $y \ge 0$. One can check by induction
$$f(x + nt) \ge (1 + t)^n f(x),$$
for all $x \in \mathbb{R}$, $t \ge 0$ and $n \in \mathbb{N}$. Let $x, z \in \mathbb{R}$, with $x < z$.
Denote $y = z - x$. For $n \in \mathbb{N}^*$ we have
$$g(z) = e^{-z}f(z) = e^{-x-y}f\left(x + n\frac{y}{n}\right) \ge e^{-x-y}\left(1 + \frac{y}{n}\right)^n f(x) = \frac{\left(1 + \frac{y}{n}\right)^n}{e^y}g(x).$$
It follows
$$g(z) \ge \lim_{n \to \infty} \frac{\left(1 + \frac{y}{n}\right)^n}{e^y} g(x) = g(x),$$
therefore $g$ is monotonously increasing on $\mathbb{R}$.