Green's representation as a classical solution of Poisson's equation If there exists Green's function $ G(x,y) $ and if we have $ u \in C^2(\Omega) $ such that $$ u(x) = \int_\Omega G(x,y)f(y)dy - 
\int_{\partial\Omega}\frac{\partial G}{\partial\nu}(x,y)g(y)dS(y) $$ then under what conditions of $\Omega, f$ and $g$ will $u$ be a solution of the problem
\begin{equation*}
 \begin{cases}
   -\Delta u = f & in\ \Omega  \\
   u = g & in\ \partial\Omega 
 \end{cases}
\end{equation*}
under usual conditions of $\Omega$ and for $ g \in C^1(\partial\Omega) $ it can be checked that $ u = g $ in $ \partial\Omega$ but how to calculate Laplacian of $u$ ?
 A: The strange thing about your question is that you ask "under what conditions" while already imposing a bunch of conditions on the problem. 
You already assume $g\in C^1(\partial \Omega)$ which is more than enough. Also, this implicitly assumes that $\partial\Omega$ is $C^1$ smooth; otherwise $C^1(\partial \Omega)$ would not make sense. Furthermore, you already assume that $u\in C^2(\Omega)$ (which is a little strange, imposing assumptions on unknown solution). This means you have $f\in C(\Omega)$, otherwise $-\Delta u=f$ is obviously impossible. 
Anyway, under the assumptions $g\in C^1(\partial \Omega)$ and $u\in C^2(\Omega)$ you have what you want: namely, $-\Delta u=f$. Since both $\Delta u$ and $f$ are  continuous,   we only have to check that $\int_\Omega (-\Delta u)\varphi = \int_\Omega f\varphi$ for every $C^\infty$ smooth compactly supported $\varphi$. The second integral in the definition of $u$ can be disregarded here: it is harmonic in $\Omega$, because $G(\cdot,y)$ is harmonic for every $y\in\partial\Omega$. For the first integral, we write 
$$
\int_\Omega (-\Delta u)(x)\varphi(x)\,dx = \int_\Omega  -u(x) \Delta \varphi(x)\,dx 
= - \int_\Omega\int_\Omega G(x,y)\Delta \varphi(x) f(y)\,dy\,dx \\ 
= - \int_\Omega f(y) \int_\Omega G(x,y)\Delta \varphi(x) \,dx \,dy
$$
The inner integral is $\varphi(y)$, since integration against $G$ reproduces smooth compactly supported functions from their Laplacian. 
