Sure one can! Just note that $f$ can be written as the "convolution" of $-g$ with an appropriate integrable kernel, the Green function $G(x)$ for the flat torus: $$f(x) = -\int_\Omega g(y) G(x - y) dy.$$ If $d \geqslant 3$, the
The Green function for a flat torus is equal todefined as follows. Pick a smooth, symmetric, compactly supported function $$G(x) = c_d \sum_{n \in 2 \mathbb Z^d} (|x+n|^{2-d} - \alpha_n),$$ where$\phi$ such that the periodisation of $\phi$ is $2^{-d}$: $$ \sum_{n \in 2 \mathbb Z^d} \phi(x + n) = 2^{-d} . $$ Let $$u(x) = c_d \biggl(|x|^{2-d} - \int_{\mathbb R^d} |x-y|^{2-d} \phi(y) dy\biggr) $$ be the solution of $-\Delta u = \delta_0 - \phi$ in full space. Then $G$ is the periodisation of $u$: $$G(x) = \sum_{n \in 2 \mathbb Z^d} (u(x + n) - \alpha_n),$$ where $\alpha_n$ is the mean value of $|x+n|^{2-d}$$u(x+n)$ over $[-1,1]^d$. If $d = 2$, replace $|\cdot|^{2-d}$ by $-\log |\cdot|$.
Indeed:
It is relatively easy to see that $u(x) \sim C_d |x|^{-d}$ as $|x| \to \infty$ (this follows by Taylor's theorem for $|x|^{2-d}$ and the symmetry of $\phi$). Hence, given any $a > 0$, the $L^1$ or $L^\infty$ norm of $v_n(x) = u(x + \alpha_n) - \alpha_n$ over $[-a,a]^d$ is of the order $|n|^{-1-d}$ as $|n| \to \infty$. Therefore, the series that defines $G(x)$ converges uniformly and in $L^1$ over $[-a,a]^d$.
Each term $u(\cdot + n)$ is harmonic in $(-a,a)^d$ when $|n|$ is large enough. Thus, by uniform convergence, we have $$ -\Delta G = \sum_{n \in 2 \mathbb Z^d} (-\Delta u(\cdot + n)) = \delta_0 - \sum_{n \in 2 \mathbb Z^d} \phi(\cdot + n) = \delta_0 - 2^{-d}$$ in $[-1,1]^d$.
A very similar argument shows that one can replace $\alpha_n$ by $\alpha_{n+2e_j}$ in the series that defines $G(x)$, and hence $G(x)$ is periodic with period $2 e_j$ for every $j = 1, \ldots, d$. Here, of course, $e_j = (0, 0, \ldots, 0, 1, 0, 0, \ldots, 0)$.
Thus, $G$ is indeed the "periodic" Green function for $-\Delta + 2^{-d}$ in $[-1,1]^d$.