I'll start from where you ended. The excited state of the atom has a finite lifetime $\tau$. Therefore, due to the energy-time uncertainty relation $\Delta E\Delta t\sim\hbar$ the energy of the excited state is not specified exactly but has a finite linewidth. Moreover the atom moves and sees the frequency of the photon Doppler-shifted and this Doppler shift is again smeared because of the uncertainty between position and momentum. The atom can also undergo collisions that lead to an exchange of energy that can channel the excess energy to the surroundings or take the missing energy from it. These are the main reasons the linewidth of the atomic transition is finite.

Mathematically, you can describe this in the perturbation theory, treating the atom-light interaction as a perturbation. Assuming only one-photon processes you can derive the Hamiltonian $H_\textrm{int} \propto a\sigma_+ +a^\dagger\sigma_-$, where $\sigma_-$ is the atomic lowering operator and $\sigma_+=\sigma_-^\dagger$. The solution of this interaction can be found in many quantum optics textbooks; it leads to oscillations of the atom between the ground and excited states. The amplitude depends on the detuning of the field and the atomic transition, i.e., the difference $\omega-\omega_0$. However, the excitation probability is finite as long as the detuning is not too large.

Finally, the absorbed photon will transfer its momentum to the atom. That is nevertheless typically much smaller than the momentum the atom has. To get the grip of it, the photon momentum is $\hbar k\sim \hbar/\lambda$ which, for visible light is of the order of $10^{-27} \textrm{kg m s}^{-1}$. If you have an atom of a gas moving typically at the speed of hundreds of meters per second, its momentum is at least to orders of magnitude higher. In a rigorous treatment of the perturbation the interaction represents, the momentum conservation can be treated more carefully.

I'll start from where you ended. The excited state of the atom has a finite lifetime $\tau$. Therefore, due to the energy-time uncertainty relation $\Delta E\Delta t\sim\hbar$ the energy of the excited state is not specified exactly but has a finite linewidth. Moreover the atom moves and sees the frequency of the photon Doppler-shifted and this Doppler shift is again smeared because of the uncertainty between position and momentum. The atom can also undergo collisions that lead to an exchange of energy that can channel the excess energy to the surroundings or take the missing energy from it. These are the main reasons the linewidth of the atomic transition is finite.

Mathematically, you can describe this in the perturbation theory, treating the atom-light interaction as a perturbation. Assuming only one-photon processes you can derive the Hamiltonian $H_\textrm{int} \propto a\sigma_+ +a^\dagger\sigma_-$, where $\sigma_-$ is the atomic lowering operator and $\sigma_+=\sigma_-^\dagger$. The solution of this interaction can be found in many quantum optics textbooks; it leads to oscillations of the atom between the ground and excited states. The amplitude depends on the detuning of the field and the atomic transition, i.e., the difference $\omega-\omega_0$. However, the excitation probability is finite as long as the detuning is not too large.

Finally, the absorbed photon will transfer its momentum to the atom. That is nevertheless typically much smaller than the momentum the atom has. To get the grip of it, the photon momentum is $\hbar k\sim \hbar/\lambda$ which, for visible light is of the order of $10^{-27} \textrm{kg m s}^{-1}$. If you have an atom of a gas moving typically at the speed of hundreds of meters per second, its momentum is at least to orders of magnitude higher. In a rigorous treatment of the perturbation the interaction represents, the momentum conservation can be treated more carefully.

EDIT

Simply put, before the interaction the atom has momentum $\mathbf{p}$. When it absorbs a photon with momentum $\hbar\mathbf{k}$ (i.e., the momentum is given by the frequency of the light, not the frequency of the atomic transition), its momentum changes to $\mathbf{p}_2 = \mathbf{p}+\hbar\mathbf{k}$. After a while it will emit another photon with (generally different) momentum $\hbar\mathbf{k}_2$ so that the resulting atomic momentum reads $\mathbf{p}_3 = \mathbf{p}_2-\hbar\mathbf{k}_2 = \mathbf{p}-\hbar(\mathbf{k}_2-\mathbf{k})$.

If you wanted to talk about the force the photon exerts on the atom, you would have to divide its momentum by the time it takes the atom to absorb the photon, not by the lifetime of the excited state. But I have never heard about anyone considering the force in such a scenario.