If you are imposing that at $x = 0$ the function $f(t,0) = 0$ for all time $t$, then necessarily that the only solution is $\Phi(t,x) = e^{-bt}c/b$. This is because your equation is a transport equation, and by prescribing data on $x = 0$ you have a well-posed evolution equation (when $a \neq 0$) where the "time variable" is $x$ and the "space variable" is $t$. And the unique solution is as given above.
But in general you probably are not looking for such a solution.
I expect what you are looking at is a problem on the first quadrant $x\geq 0$ and $t\geq 0$, with boundary condition prescribed at $x = 0$ when $t \geq 0$ and with initial condition prescribed at $t = 0$ for $x \geq 0$.
Then what we have is that $f(t,0)|_{t\geq 0} = 0 \implies \frac{c}{b} e^{-bt} = \Phi(-at)|_{t\geq 0}$ This only gives the data for $\Phi(s)$ when $s \leq 0$. You are still free to prescribe $\Phi(s)$ for $s > 0$ using the initial condition $f(0,x)|_{x\geq 0}$. That is $ \Phi(s)|_{s\geq 0} = \left(f(0,s) + \frac{c}{b}\right)e^{-bt}$
Note that when $a = 0$ the boundary condition only tells you that $\Phi(0) = c/b$. It tells you absolutely nothing about $\Phi$ for other values of the argument.
Edit after original question was updated
If you are now prescribing initial-boundary data, where $f(t,0) = 0$ for positive times and $f(0,x) = 0$ for positive $x$, then what we need to do is to see how this translates to conditions on $\Phi$. What you need to pay attention to is the fact that the initial and boundary data can only be used to prescribe the free function on corresponding domains. See below for an illustration.
Now $f(t,x) = - \frac{c}{b} + e^{bt}\Phi(x - At^2 / 2)$. We solve this to get $ \Phi(x - At^2/2) = e^{-bt}\left(f(t,x) + \frac{c}b \right) $
Using that if $t > 0$ and $x = 0$, $f(t,x) = 0$. We have that
$ \Phi(0 - At^2/2) = e^{-bt}\frac{c}b $
for every $t > 0$. This tell you that, changing variables via $s = At^2/2$, we have that for every $s > 0$ (solving $t = \sqrt{2s/A}$)
$ \Phi(-s) = e^{-b \sqrt{2s/A}} \frac{c}{b} $
This specifies your arbitrary function along the negative real line.
Now we use that if $t = 0$ and $x > 0$, $f(t,x) = 0$. Plugging in we have
$ \Phi(x - 0) = e^{-b\cdot 0} \frac{c}{b} = \frac{c}{b} $
this means that $\Phi(x)$ for $x$ positive is equal to $c/b$.
In other words, you've almost gotten the answer already! What you need to pay attention to is the domain over which your initial/boundary can be applied to give you information. In this case, the boundary data can only be used to specify $\Phi$ along the negative real line, while the initial data can only be used to specify $\Phi$ along the positive real line.