How does this then imply that the solution is $u = f(y^\prime) = f(bx - ay)$?

Solution 1:

After the coordinate transformation, we know that $u=f(x',y')$. But since $u_{x'}=0$, we get that for a fixed $y'$ and variable $x'$, $f(x',y')$ is a constant. In other words, the value of $f(x',y')$ depends only on $y'$. With an abuse of notation, we can thus write $f(x',y')=f(y')=f(bx-ay)$.

If we wanted to be a bit more clear on the notation, we could choose an arbitrary $x'$, say $x'=0$, and then define the function $\tilde f(y')=f(0,y')$. Since $f$ was an arbitrary function of $(x',y')$, we get $\tilde f$ is an arbitrary function of $y'$, that is, $\tilde f$ is an arbitrary function of $bx-ay$.