I think it is as easy as this: if \(f\) has a pole at \(z_0\) then \(\frac{1}{f}=:g\) is analytic there, and then we have
\[ f^{-1}(c^t f(z)) = g^{-1}\left(\frac{1}{c^t\frac{1}{g(z)}}\right) = g^{-1}(c^{-t} g(z)) \]
and hence it is analytic at the fixed point.
A similar reasoning works with the Abel function
\[ f^{-1}(t+ f(z)) = g^{-1}\left(\frac{1}{t+\frac{1}{g(z)}}\right) = g^{-1}\left(\frac{ g(z)}{tg(z)+1}\right) \]
\[ f^{-1}(c^t f(z)) = g^{-1}\left(\frac{1}{c^t\frac{1}{g(z)}}\right) = g^{-1}(c^{-t} g(z)) \]
and hence it is analytic at the fixed point.
A similar reasoning works with the Abel function
\[ f^{-1}(t+ f(z)) = g^{-1}\left(\frac{1}{t+\frac{1}{g(z)}}\right) = g^{-1}\left(\frac{ g(z)}{tg(z)+1}\right) \]
