It is well known that if $f\in \mathcal{D}'(\mathbb{R}^3,\mathbb{R}^3)$ and $\textbf{curl} f= 0$ then there exists a $u\in \mathcal{D}'(\mathbb{R}^3)$ such that $\nabla u = f$.
Question. Does the same result still hold with $f\in \mathcal{S}'(\mathbb{R}^3,\mathbb{R}^3)$ and $u\in \mathcal{S}'(\mathbb{R}^3)$?
As reuns commented, $\mathcal S'$ is closed under Fourier transforms.
Using $\hat f(k) = \int_{\mathbb R^3} f(x) \, e^{-i k \cdot x} \, dx$ the Fourier transform of $\nabla \times f(x) = 0$ becomes $ik \times \hat f(k) = 0$. Thus $\hat f(k)$ is parallel to $ik$ for all $k \in \mathbb R^3,$ i.e. there is a function $\hat u \in \mathcal S'(\mathbb R^3, \mathbb R)$ such that $\hat f(k) = ik \, \hat u(k).$ An inverse Fourier transform then gives $f(x) = \nabla u(x).$