The key difference is that Compton scattering occurs on a free electron. It's all about making energy conservation work out. In an atomic process, the atom goes form having energy $E_1$ to $E_2$, and its energy is fixed by the atomic state it goes to. It can't be a $2 p ^1$ electron and also have extra energy it hangs onto. (Things are slightly more complicated than this, due to splitting of energy levels, Doppler shifting, etc. But this still just means you have a slightly broader window, rather than a true single-infinite-precision-frequency-condition.) So the photon has to have $E_2 - E_1$.
On the other hand, a Compton-scattered photon hits an electron with, say, no energy (for simplicity), which then goes off into having any number of energies. It's only limited by how much energy the photon had. Other than that, there's no reason that after collision it can't be a 20 $keV$ electron, or a 30 $keV$ electron, or anything. So there isn't the same sharp absorbance peak, because the energy condition is less stringent.
Incidentally, something like this can happen in atomic process if the photon is high enough energy to eject the electron from the atom entirely. In this case, we say the electron is excited "to the continuum"--reflecting the fact that once the electron is free, it has a whole continuum of available energies, instead of just the few bound-state energies allowed by the atomic potential.
EDIT: I just realized I ended up answering some questions raised in the comments and not the asked question. My point here is to agree with others that the photon is best viewed as having been absorbed and re-emitted in both processes, and to explain why the energetic considerations are different in each.