I got stuck on part of a question, and I have the feeling there's a simple approach that I'm not thinking of. Here is my work so far:
Ultimately, we would like to solve the PDE
$$u_t+cu_x = u\tag{T}$$ $$u(x,0) = f(x) \tag{IC}$$
With the substitution $u(x,t) = v(x - ct,t)$, we get the modified version
$$v_t = u = v(x-ct,t)$$ $$v(x,0) = f(x)$$
I think that there should be a way of finding a general solution to that first equation $$v_t = v(x-ct,t)$$ Certainly, $Ce^t$ would be a valid solution. However, the general solution must somehow involve an undetermined function of $x$, and I'm not sure how to bring it in to here. My immediate thoughts were to try $\psi(x)e^t$ and $\psi(x-ct)e^t$, but the first never works, and the second will not work for an undetermined differentiable function $\psi$. At this point, I feel as though I've run out of ideas.
Any help or hints are appreciated.
(I hope my notation is not too horrendous) We used the method of characteristics to solve this equation in my PDEs class. The characteristic equations are given by, $$\frac{\partial{t}}{\partial\tau}=1,\quad\frac{\partial x}{\partial \tau}=c,\quad\text{and}\quad \frac{\partial u}{\partial \tau}=u $$ where $\tau$ is a parameter. I get $t=\tau$, $x=c\tau+x(0)$, and $u=u(x(0),0)e^\tau$. (Here $x(0)$ is just my initial value when our parameter $\tau=0$.) The initial conditions give $$ u=f(x(0))e^t $$ Solving for $x(0)$ gives us that $u(x,t)=f(x-ct)e^t$, which is essentially what you had. Taking partial derivatives, we see this solution satisfies our PDE. Hope that helps!