This is known as the Goursat problem, because the boundary condition are given over two intersecting characteristic lines. Notice the necessary condition (which turns out to be sufficient):
$$c_1(0)=c_2(0).$$
There is no closed form in general, but I can explain here the proof of existence and uniqueness, which gives also the regularity of the solution. Amazingly, the proof mimics that for an ODE (usually, proffs for PDEs are much different than for ODEs). You must rewrite the Goursat problem as a fixed point question, in terms of the unknown $(u,v:=\partial_y u)$ for instance. we have
$$u(x,y)=c_2(x)+\int_0^yv(x,\eta)d\eta$$
and
$$v(x,y)=c_1'(y)+\int_0^x(Au)(\xi,y)d\xi.$$
Therefore $(u,v)$ is the fixed point of the map $T:(u,v)\mapsto (U,V)$ defined by
\begin{eqnarray*}
U(x,y) & = & c_2(x)+\int_0^yv(x,\eta)d\eta, \\
V(x,y) & = & c_1'(y)+\int_0^x(Au)(\xi,y)d\xi.
\end{eqnarray*}
Exactly as in the case of ODEs, $T$ is contracting over $C^k\times C^{k-1}([-a,a]\times[-a, a])$ for some finite $a$, provided that $c_{1,2}\in C^k$ and $A\in C^k$. Hence it has a unique fixed point. And $a$ is known in terms of $A$ only. Given a bounded rectangle $R$, you can cover $R$ with overlaping squares of side $2a$, and you get a unique solution on $R$.
Remark. On the qualitative level, the Goursat problem has an amazing property: suppose for instance that $c_1\equiv0$ over ${\mathbb R}$, and that $c_2\equiv0$ over $(-\infty,0]$. Then
$$(x<0)\Longrightarrow(u(x,y)=0).$$
In my paper La compacité par compensation pour les systèmes hyperboliques non linéaires de deux équations à une dimension d'espace (JMPA 1986), I used these half-plane-supported solutions, which I called of east-type. There are of course solutions of north-, west- and south-types.
Equivalently, the value of $u$ in the quadrant $x,y>0$ depends only upon the restrictions of $c_1$ on $(0,y)$ and of $c_2$ on $(0,x)$. In the limit as $y\to0$, we obtain a closed formula for the normal derivative
$$\partial_yu(x,0)=\int_0^xA(\xi,0)c_2(\xi)d\xi.$$
