The PDE I'm trying to solve is the following : $$(1-xu)u_x-(1-yu)u_y=(y-x)u$$
Using the method of characteristics we have: $$x'(t)=1-xu$$ $$y'(t)=yu-1$$ $$u'(t)=(y-x)u=yu-xu$$
A first relationship can easily be derived this way :
$$u'(t)-x'(t)-y'(t)=0\space\space\space(1)$$ $$=>u(t)-x(t)-y(t)=C_1$$
It took me quite some time to get the second relationship in this problem but here's what I found dividing $x'(t)$ with $y'(t)$ $$yux'(t)-x'(t)=y'(t)-xuy'(t)$$ $$=>u[yx'(t)+xy'(t)]=x'(t)+y'(t)$$ $$(1)=>u[yx'(t)+xy'(t)]=u'(t)$$ $$yx=\ln|u|+C_2$$ $$u=e^{yx}+C_2$$
Since $C_2$ will be a function of $C_1$ we get this implicit form: $$u=e^{yx}+f(C_1)$$ $$u=e^{yx}+f(x+y-u)$$ Is it a problem that u is in the arbitrary function? And if not, how do I apply this result in the PDE to check if it is satisfied?
There is a little mistake in your calculus : $$yx=\ln|u|+C_2\quad\text{is OK.}$$ $$u=e^{yx}+C_2\quad\text{is false. It should be :}$$ $$u=c_2\:e^{yx}$$ where $c_2=e^{-C_2}$
As a consequence, the general solution of the PDE, expressed on the form of implicit equation, is : $$u=e^{xy}f(x+y-u) \tag 1$$ Without boundary condition, it is not possible to transform the implicit equation into an explicit form.
The function $f$ has to be determined according to some boundary condition(s). Depending on the boundary condition, the solution $u(x,y)$ can (or not) be expressed explicitly.
Nevertheless, one can put this result $(1)$ into the PDE and check if it is satisfied. Compute the partial derivatives such as :
$u_x=ye^{xy}f+e^{xy}f'-e^{xy}u_xf'\quad\text{leading to}\quad u_x=\frac{yf+f'}{e^{-xy}+f'}\tag 2$
$u_y=xe^{xy}f+e^{xy}f'-e^{xy}u_yf'\quad\text{leading to}\quad u_y=\frac{xf+f'}{e^{-xy}+f'}\tag 3$
Then, put $u$ from Eq.$(1)$ , $u_x$ from Eq.$(2)$ and $u_y$ from Eq.$(3)$ into the PDE. Simplify. You will see that it is satisfied.