I think you are misinterpreting the statement that "it doesn't have any effect". This statement doesn't mean that the Faddeev-Popov methodology "doesn't work", as you wrote later. Instead, it means that it is completely unnecessary.
If you look at the Faddeev-Popov ghosts' Lagrangian, you will see that for Abelian groups, the structure constants $f_{abc}$ vanish and we're left with
$$ {\mathcal L}_{\rm ghost} = \partial_\mu \bar c^a \partial^\mu c^a $$
which means that the ghosts are completely decoupled. They don't interact with the gauge fields (photons). You may still use the Faddeev-Popov machinery and the BRST formalism based upon it to identify the physical states as the cohomologies of $Q$, the BRST operator.
But what this BRST machinery tells you is something you may easily describe without any Faddeev-Popov ghosts, too. It just tells you that the excitations of $\bar c,c$ are unphysical much like the excitations of time-like and longitudinal photons. That's why the BRST problem in the case of Abelinan gauge groups is "solvable" in such a way that you may simply eliminate the ghosts completely, together with 2 unphysical polarizations of the photon. And that's why QED may be taught without any Faddeev-Popov ghosts and one may still construct nice Feynman rules for any multiloop diagrams.
For non-Abelian theories, the counting still works – ghosts, antighosts, and two polarizations of gluons etc. are unphysical. However, because there are interactions of ghosts with the gluons in that case, there's no easy way to describe the physical states without the Faddeev-Popov ghosts.
