Let $\Omega\subset\Bbb R^n$ be a domain (i.e. some open subset with smooth boundary). Recall that we can define the space of distributions $\mathcal{D}'(\Omega)$ as the topological dual of $C^\infty_{\mathrm c}(\Omega)$. Furthermore, for any distribution $u\in\mathcal{D}'(\Omega)$, and for $1\le p\le\infty$, we can define the $L^p(\Omega)$ norm $$\lVert u\rVert_{L^p(\Omega)}=\sup_{\lVert\varphi\rVert_{p'}=1}\lvert\langle u,\varphi\rangle\rvert$$ which corresponds with the usual $L^p$ norm whenever $u=f\in L^p$.
Now, if $\lVert u\rVert_{L^p(\Omega)}<\infty$, then $u$ can be extended (using Hahn-Banach) to some bounded linear functional $\ell\in L^{p'}(\Omega)'$. If $p>1$, then we know that $\ell=\langle f,\cdot\rangle$ for some $f\in L^p(\Omega)$, and since $C_{\mathrm c}^\infty(\Omega)\subset L^{p'}(\Omega)$, we have that $u=f$ as distributions.
But what if $\lVert u\rVert_{L^1(\Omega)}<\infty$? Then must we have $u\in L^1$?
As user254433 pointed out, having finite $L^1$ norm in the distributional sense does not imply we have an $L^1$ function -- it could be a singular finite measure instead, such as the Dirac delta.
It is difficult to characterize $L^1$ functions through their actions on some other things, because $L^1$ is not the dual space of any normed space.
One can describe $L^1$ functions as measures that are absolutely continuous with respect to the Lebesgue measure $dx$. This leads to the characterization that user254433 suggested: given an open set $\Omega$, define $$ \|u\|_{1,\Omega} = \sup \{|\langle u,\varphi\rangle|\colon |\varphi|\le 1, \ \operatorname{supp} \varphi\subset \Omega \} $$ be the "restricted $L^1$ norm of $u$". Then $u\in L^1$ if and only if for every $\epsilon>0$ there exists $\delta>0$ such that $\|u\|_{1,\Omega}<\epsilon$ whenever the Lebesgue measure of $\Omega$ is less than $\delta$.
This isn't really saying much; it's just the absolute continuity of a measure expressed in terms of a distribution.