Master Theorem on a*T(n/b) + c?
Yes. The constant function $c$ plays the part of $f(n)$. Stated differently, you can use the function $f(n) = c$ to apply the master theorem.

A perhaps simpler way is to find a shift $d$ such that $S(n) := T(n) + d$ satisfies a nicer recurrence. We have
$$
S(n) = T(n) + d = aT(n/b) + c + d = a(S(n/b)-d)+c+d = aS(n/b) +c+d-ad.
$$
If $c+d-ad = 0$, then we get the simple recurrence $S(n) = aS(n/b)$, whose solution is $S(n) = n^{\log_b a} S(1)$ (for powers of $b$). This happens when $d = c/(a-1)$. We conclude that $T(n) = n^{\log_b a}(T(1) + c/(a-1)) - c/(a-1)$ (for powers of $b$).