The answer is yes. Here is a sketch of the argument.
Claim 1: Suppose that $u \in L^1$ and $\Delta u \in \mathcal M$. Let $f = u - \Delta u$. Then $f \in \mathcal M$ and $u = \mathcal B_2 * v$, where $\mathcal{B}_\alpha$ is the Bessel potential kernel.
Formally, this is clear, as $\mathcal{B}_2$ is the inverse of $(\operatorname{Id} - \Delta)$. I did not attempt to write a rigorous proof, but this should not be very difficult, using, for example, the ideas from Stein's Singular integrals and differentiability properties of functions.
Claim 2: The gradient of $\mathcal B_2$ is integrable.
This follows easily from the explicit expression for $\mathcal B_2$.
It follows that $$\|\nabla u\|_1 = \|\nabla \mathcal (B_2 * f)\|_1 = \|(\nabla \mathcal B_2) * f\|_1 \leqslant \|\nabla \mathcal B_2\|_1 \|f\|_{\mathcal M} < \infty,$$ as desired.