More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$.
The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling between the em. field and electrons is rather weak), so a process where two photons interact is "unlikely"$^2$ and thus strongly suppressed. So for small intensities the linear process will dominate distinctly. The question is only, at what intensities the second order effects will become visible.
In the paper by Richard L. Smith, "Two-Photon Photoelectric Effect", Phys. Rev. 128, 2225 (1962) the photocurrent for radiation above half of the cutoff frequency but below the cutoff frequency is discussed. They note that for usual intensities the photocurrent will be minute, but that given strong enough fields
such as those observed in a focus spot of a laser (on the order of $10^7\,\mathrm{W/cm^2}$) the effect might be measurable. They also note, that thermal heating by the laser field may make the pure second order effect unobservable.
The more recent paper S. Varró, E. Elotzky, "The multiphoton photo-effect and
harmonic generation at
metal surfaces", J. Phys. D: Appl. Phys. 30, 3071 (1997) dicusses the case where high intensities (on the scale of $10^{10}\,\mathrm{W/cm^2}$) produce even higher order effects (and unexpectedly high, coherent non-linear effects, that is absorption of more than two photons by one electron). Their calculations explain the experimental observations of sharp features in the emission spectra of metal surfaces.
Historical fun fact: The 1962 paper is so old, that it talks about an "optical ruby maser"; lasers where so new back then, they did not even have their name yet.
An atom is nothing but a bounded state of electrons and a positively charged core called nucleus. The electrons in the atom are in bound state and so their energy levels are quantized. Also, it is possible to have quantized rotational and vibrational energy levels of the molecules. The way in which they differ is in the difference in the energy characterizing the transition from one state to another.
Possible ways in which a photon is absorbed by an atom or a molecule
If the energy level of the incoming photon is such that the electrons can have a transition from a state to some higher permissible state, then the photon energy level will be in the visible or ultraviolet range and we make use of this principle in electronic spectroscopy.
Suppose, a particular electron is in the energy state with energy eigenvalue $E_i$. There exists a higher energy level $E_f$. If the energy levels of the electron bound states are such that it precisely matches with the energy of the photon: $h\nu=E_f-E_i$, then the electron will get excited to the energy state $E_f$.
Now, if the incident photon energy matches the difference in the vibrational energy levels of any pair of states of the molecule, then it can cause transition from that vibrational energy state to the higher energy state. This energy usually lies in the infrared region and the technique is used in infrared spectroscopy.
For example, in the case of diatomic molecules, the vibrational energy levels are quantized and in a good sense they can be approximated to that of a harmonic oscillator: $E_n=\left(n+\frac{1}{2}\right)\bar{h}\omega$. So, if the photon energy is such that $h\nu=E_f-E_i$, the electron transits from the state $E_i$ to $E_f$, where $E_i$ and $E_f$ are given by the above equation of the harmonic oscillator and the states are defined by the quantum number $n=i$ and $n=f$.
Now, if the absorption of a photon can only affect the rotational energy levels of the molecule, then the absorbed photon will be in the microwave region. The spectroscopic technique making use of this principle is the microwave spectroscopy.
For example, the rotational energy levels of a diatomic molecule are given by: $\displaystyle{E_j=\frac{j(j+1){\bar{h}}^2}{2I}}$, where $I$ is the moment of inertia and $j$ is the angular momentum quantum number. In such a case, we can write: $h\nu=E_f-E_i$ and the bound state absorbs the photon and will get excited to the state with energy $E_f$, with $E_f$ and $E_i$ determined by the quantum number $j=f$ and $j=i$.
Now, the energy can be absorbed by the nuclei also. It can be elastic nuclei scattering (analog to very low energy Compton scattering by an electron. In this process, a photon interacts with a nucleon in such a manner that a photon is re-emitted with the same energy), inelastic nuclei scattering (the nucleus is raised to an excited level by absorbing a photon. The excited nucleus subsequently de-excites by emitting a photon of equal or lower energy) and Delbruck scattering (the phenomenon of photon scattering by the Coulomb field of a nucleus, also called nuclear potential scattering, which can be thought of as virtual pair production in the field of the nucleus. i.e., pair production followed by annihilation of the created pair). However, these processes are negligible in photon interactions.
Conclusion:
Absorption of a photon will occur only when the quantum energy of the photon precisely matches the energy gap between the initial and final states of the system. (the atom or a molecule as a whole) i.e., by the absorption of a photon, the system could access to some higher permissible quantum mechanical energy state. If there is no pair of energy states such that the photon energy can elevate the system from the lower to the upper energy state, then the matter will be transparent to that radiation.
So, if any of the above types of energy transition take place, that will affect the quantum state of the system as a whole (transits the system from one state to another). So one could say, as @annav pointed out, it is the atom (or the molecule) that absorbs the radiation and changes the energy levels of its constituent particles, depending on the energy absorbed. Anyway, a change in energy level of the electron, or rotational or vibrational energy levels of the molecules can be seen as changing the quantum state of the molecule. So, it's better to stick with the concept of the molecule as a whole absorbs the energy and changes its state to some higher energy state by changing the quantum state of its constituent particles.
Best Answer
Let us talk of free atoms, gas.
If the atom is ionized, there will be an available energy level that an electron could occupy. A free floating electron at rest relatively to the atom can fall on that energy level and release a photon. In the case of an ionized hydrogen atom ( called a proton),
it will release a photon of energy 13.6 eV .
If the electron is not at rest with the nucleus, the probability of capture is very low, though computable, the excess energy released in the interaction as a photon carrying away the difference and bringing it at rest so as to be captured. The probability is low because extra electromagnetic vertices will be needed to compute the interaction crossection.
So the answer is that predominantly the electron must be at rest to be captured.
I read this as : in a neutral atom where the electron is bound , as with the hydrogen example above, in the ground state , the electron can be kicked out of the ground state if the energy is equal or larger than 13.6 eV. The electron will carry the balance of the energy of the photon.
Excitation means that the electron is hanging around in the higher than ground state levels, it is still bound by the potential of the nucleus. Ionization is where the electron becomes free of the potential of the nucleus, and since over the 13.6 eV in the example above, there exists a contiuum, it can carry all the left over energy as kinetic energy from the kick.