Suppose $u\in L^2(\Omega)$, then for any $\phi\in C_c^\infty(\Omega)$ we have $$ \int_\Omega v\,\phi\,dx=\int_\Omega u\Delta \phi\,dx $$ Then can I conclude that $u\in H_0^1\cap H^2(\Omega)$ and $$\Delta u=v $$
Also assume that $\Omega\subset\mathbb R^N$ is open bounded, smooth boundary.
Integrating by part, we can get $$\int_\Omega u\frac{\partial^2\phi}{\partial x_i^2}\,dx=-\int_\Omega \frac{\partial u}{\partial x_i}\frac{\partial\phi}{\partial x_i}\,dx+\int_{\partial \Omega}u \frac{\partial\phi}{\partial x_i}\nu_i\,dS$$ and $$\int_\Omega \frac{\partial u}{\partial x_i}\frac{\partial\phi}{\partial x_i}\,dx=-\int_\Omega \frac{\partial^2 u}{\partial x_i^2}\phi\,dx+\int_{\partial \Omega}\phi \frac{\partial u}{\partial x_i}\nu_i\,dS.$$
If $u\in H_0^1(\Omega)$ and $\phi\in C_c^\infty(\Omega)$, then $$\int_\Omega (v-\Delta u)\phi\,dx=0,$$ which implies $$\Delta u=v.$$ If $u\in L^1(\Omega)$ only, we can not get $\Delta u=v$ from $\int_{\Omega}v\phi\,dx=\int_{\Omega}u\Delta \phi\,dx$, because boundary terms during the integration would remain. In order to obtain $u\in H^2$, the condition $v\in L^2(\Omega)$ and $\partial \Omega\in C^2$ should be hold.