Let $v(x) = |x|^{2-N}$, let $a = (1, 0, \ldots, 0)$ be a unit vector, let $$\Delta_a f(x) = f(x + a) - f(x)$$ be the difference operator, and finally let $u$ be the third-order finite difference of $v$: $$u(x) = \Delta_a^3 v(x) .$$ Then:

• $\Delta u$ is a finite measure concentrated at $\{0, a, 2a, 3a\}$;

• $u$ is locally integrable, $u \approx |x|^{-1-N}$ at infinity, and hence $u$ is integrable;

• $|\nabla u| \approx |x|^{-N}$ at infinity, and hence $\nabla u$ is not integrable.

Thus, $u$ is a counter-example.

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.