Perhaps, I'm reading the problem statement wrong, and it's not asking for existence, only uniqueness; but in any case...
Problem. Let $g\in L^{1}(\mathbb{R}^{n})$, $\|g\|_{L^{1}}<1$. Prove that there is a unique $f\in L^{1}(\mathbb{R}^{n})$ such that $$f(x)+(f\ast g)(x)=e^{-|x|^{2}}, \quad x\in\mathbb{R}^{n} \enspace\text{a.e.}$$
The uniqueness isn't hard to show. Suppose such an $f\in L^{1}(\mathbb{R}^{n})$ exists. Then taking Fourier transforms under the convention
$$\widehat{f}(\xi)=\int_{\mathbb{R}^{n}}f(x)e^{-2\pi i x\cdot\xi}dx,\quad\xi\in\mathbb{R}^{n}$$
we obtain
$$\widehat{f}(\xi)+\widehat{f}(\xi)\widehat{g}(\xi)=Ce^{-c|\xi|^{2}}$$
where $C,c$ are some constants that don't matter. Since $\|\widehat{g}\|_{\infty}\leq\|g\|_{L^{1}}<1$, $(1+\widehat{g}(\xi))^{-1}$ is a bounded, positive continuous function. Hence, we can write
$$\widehat{f}(\xi)=C\frac{e^{-c|\xi|^{2}}}{1+\widehat{g}(\xi)},\quad\forall \xi\in\mathbb{R}^{n} \tag{*}$$
Fourier inversion then completes the proof of uniqueness.
Regarding existence, since the RHS of * is rapidly decaying, we can take the inverse Fourier transform of both sides of * to obtain a $C_{0}$ $f$. And by approximating $g$ by Schwartz functions, we probably get by distribution theory that $f$ satisfies the functional equation (the convolution is well-defined since $f\in L^{\infty}$). But since the only regularity for $\widehat{f}$ is continuity, it's not clear to me that $f\in L^{1}(\mathbb{R}^{n})$. Any thoughts?
Hint: Note that $$\frac{1}{1+\hat g}=1-\hat g+\hat g^2\dots.$$ Since $||g||_1<1$, the series $$h=-g+g*g-g*g*g...$$converges in $L^1$.
This is actually a special case of something about linear operators: If $||T||<1$ then $I-T$ is invertible. (Which is a special case of something about Banach algebras...)