I need to solve the following Cauchy problem: $$\left\{\begin{matrix}xu_x-yu_y = u\\ u(x,0) = h(x)\end{matrix}\right.$$
What we know so far is the method of characteristics, so here we go:
Define $X(s,t) = (x(s,t),y(s,t),z(s,t))$, we want then to solve the problem $$\left\{\begin{matrix}\frac{dX(s,t)}{dt} &= (x(s,t),-y(s,t),z(s,t))\\ X(s,0) &= (s,0,h(s))\end{matrix}\right.$$
if the functions $x(s,t),y(s,t)$ are invertible, we will find the solution $u(x,y) = z(s(x,y),t(x,y))$ which satisfies our problem. We then have:
$\frac{dx(s,t)}{dt} = x(s,t) \implies x(s,t) = e^tk_1(s)\, \mbox{ given that }x(s,0) = s \implies x(s,t) = e^ts$ $\frac{dy(s,t)}{dt} = -y(s,t) \implies y(s,t) = e^{-t}k_2(s)\, \mbox{ given that }y(s,0) = 0 \implies y(s,t) = 0\, \forall s,t$${\color{red}{\mbox{ Here's the problem}}}$ $\frac{dz(s,t)}{dt} = z(s,t) \implies z(s,t) = e^tk_3(s)\, \mbox{ given that }z(s,0) = h(s) \implies z(s,t) = e^th(s)$
As the bondary values are given with $y=0$, the field vector defined by this PDE has component $0$ in the $y$ direction, so any solutions passing by the hiperplane $y=0$ must stay there. So I'm actually solving the ODE
$$\left\{\begin{matrix}xu_x = u\\ u(x) = h(x)\end{matrix}\right.$$ which we can easily solve by $\frac{u_x}{u} = \frac{1}{x} \implies \ln(u(x)) = \ln(x)+C \implies u(x) = C_2x.$
Therefore our problem only has solutions if the function $h(x) = Kx$ for some constant $K$, and that solution would be $u(x,y) = Kx \;\forall y$.
Is this correct?
I would say the problem is ill-posed. The first difficulty is you didn't say where the equation holds. Is $x u_{x} - yu_{y} = u$ supposed to hold in $\{(x,y) \in \mathbb{R}^{2} \, \mid \, y > 0\}$? Where is the "action" taking place?
The real difficulty is the boundary $y = 0$ is "characteristic." As you noticed, any characteristic that starts on this line stays there for all time. Hence the information about $u$ on the line doesn't propagate anywhere --- so it tells me nothing about $u$ elsewhere in the domain. (Hence part of the reason why you should specify where the equation holds...)
It could be an interesting exercise to try to find a function $h$ for which there's more than one solution to the PDE. However, again, you need to specify where the equation holds.