First, unless I misinterpret what you wrote, your integral representation for $u(t,x)$ is not correct. At the very least, it's missing boundary terms. Without boundary terms the formula obviously fails for any $u(t,x)\ne 0$ with $Lu(t,x)=0$. Whatever the right boundary term is, that will give you the $L$-analog of the Poisson kernel.

Here's how to find it. First, you need the formal adjoint $L^*$ of the operator $L$. Everybody knows how to find the formal adjoint, by using integration by parts. But what that really implies is that there exists a pair of bidifferential operators $V[-,-]$ and $W[-,-]$, linear in each argument such that
$$
v L[u] - L^*[v] u = \partial_t V[v,u] + \operatorname{div}_x W[v,u] .
$$
Basically, $V$ and $W$ collect all the boundary terms that you get by moving all the derivatives from $u$ to $v$ using integration by parts. If that's confusing, see this recent answer for some explanation. In your case,
\begin{align*}
L&=\frac{\partial}{\partial t}+\operatorname{div}_x\big(A(x,t)\nabla_x\big) , \\
L^*&=-\frac{\partial}{\partial t}+\operatorname{div}_x\big(A(x,t)\nabla_x\big) , \\
V[v,u] &= v u , \\
W[v,u] &= v A(t,x) \nabla_x u - u A(t,x) \nabla_x v .
\end{align*}

Next, you actually need $G(s,y;t,x)$ to satisfy
$$
L_{s,y}^*[G](s,y;t,x) = \delta(s-t) \delta(y-x) .
$$
Then, for $(t,x) \in \Omega = [0,T] \times D$, and assuming that $G$ and $u$ are sufficiently regular for all the differentiations to be well defined (perhaps in a weak sense),
\begin{align}
u(t,x)
&= \int_\Omega \delta(s-t) \delta(y-x) u(s,y) \, ds dy \\
&= \int_\Omega L_{s,y}^*[G](s,y; t,x) u(s,y) \, ds dy \\
&= \int_\Omega \left( \partial_s V[u,v] + \operatorname{div}_y W[u,v] \right) \, ds dy
+ \int_\Omega G(s,y; t,x) L_{s,y}[u](s,y) \, ds dy \\
&= \left. \int_D V[G,u] \right|_{s=0}^T \, dy
+ \int_{\partial D}\left( \int_0^T n(y)\cdot W[G,u] \, ds \right) dy_{\|}
+ \int_\Omega G(s,y; t,x) L_{s,y}[u](s,y) \, ds dy ,
\end{align}
where $n(y)$ is the outer normal vector to the boundary $\partial D$ and $dy_{\|}$ is the surface area element of $\partial D$.

So, in your case, the final integral representation of $u(t,x)$ inside $\Omega$ is
\begin{multline*}
u(t,x) = \left. \int_D G(s,y; t,x) u(s,y) \right|_{s=0}^T \, dy \\
+ \int_{\partial D} \int_0^T \left( G(s,y; t,x) \, n(y)\cdot A(s,y)\nabla_y u(s,y) - u(s,y)\, n(y)\cdot A(s,y) \nabla_y G(s,y; t,x) \right) \, ds dy_{\|} \\
+ \int_\Omega G(s,y; t,x) L[u](s,y) \, ds dy .
\end{multline*}
Depending on what properties you require of $u$ and $G$ in the interior and on the boundary of $\Omega$, some of the terms on the right hand side of that identity will vanish. For instance, if $L[u] = 0$, then you are left with a boundary integral representation of the form
$$
u(t,x) = \int_{\partial\Omega} p(s,y; t,x) \, (ds dy)_{\|} ,
$$
where again the $\|$ subscript denotes the appropriate surface area element on $\partial \Omega$. I guess $p(s,y; t,x)$ would be your $L$-analog of a Poisson kernel, which you can read off from the preceding identity.