Assume $a,b,c$ arbitrary variables or real numbers.
For $au_x+bu_y=0$ we have that the solution is $u(x,y)=f(bx-ay)$ by the Geometric Method,
For a differential equation $u_x+yu_y=0$, we may reduce to $\dfrac{dy}{dx}=\dfrac{y}{1}$ and obtain a solution of the form $u(x,y)=f(e^{-x}y)$
How to solve $au_x+bu_y+cu=0$ with either Geometric Method or reduction of PDE to ODE?
We have $$\frac{dx}{a} =\frac{dy}{b} =\frac{du}{-cu}$$ Hence $$bx-ay=k_1 , u=k_2e^{\frac{-c}{b} y}$$ therefore the solution is $$u(x,y) =e^{\frac{-c}{b} y} f(bx-ay)$$
Using the method of characteristics we have
$$\frac{dx}{a}=\frac{dy}{b}=-\frac{du}{cu}$$
So we have
$$\frac{dy}{dx}=\frac{b}{a}\implies bx-ay=c_1$$
and $$\frac{du}{dx}=-\frac{cu}{a}\implies u=c_2e^{-\int\frac{c}{a}dx}=c_2e^{-\frac{c}{a}x}\implies ue^{\frac{c}{a}x}=c_2$$
and combining forms we obtain the general solution
$$u(x,y)=e^{-\frac{c}{a}x}f(bx-ay)$$
for some arbitrary function $f$.
Another method mentioned by EditPiAf in the comments here is to set $v=ue^{\frac{c}{a}x}$ and substitute into the PDE which gives
$$av_{x}+bv_{y}=au_xe^{\frac{c}{a}x}+cue^{\frac{c}{a}x}+bu_{y}e^{\frac{c}{a}x}=\left(au_{x}+bu_{y}+cu\right)e^{\frac{c}{a}x}=0$$ $$\implies v(x,y)=f(bx-ay)$$ $$\implies u(x,y)=e^{-\frac{b}{a}x}f(bx-ay)$$
