Correct. [Old photon energy level] - [electron energy level jump] = [new photon energy level]
HUH!?!?!? There's a thought here that can be manipulated to be correct, but I think you might want to use the right terminology. There is no "new photon energy level" or "old photon energy level", as a photon simply has energy. Electrons exist in energy levels, not photons. Photons are absorbed during electronic excitation and emitted during electronic relaxation. The photon involved in the process has an energy equal to the transition energy. So you could make your equation correct by rewriting as follows.
deltaE = hf = Final electronic energy level - Initial electronic energy level
Because energy levels are quantized, the gaps between levels (transition energies) are also quantized. You absorb a photon with lamda = 412 nm only if the energy of that photon matches the energy difference between the initial energy level (ground state) and the final energy level (excited state). For that specific transition, 411 nm won't work as 413 nm won't work; only the exact wavelength works. They may work for another transition, but only one photon works for a specific transition.
When Econ is talking about "kinetic energy to the electron", he means the case where an electron is knocked completely out of the atomic shell. In that case, instead of emitting a photon, the electron retains the "extra" energy as kinetic energy. But if you're talking electron shell transfer, the "extra" energy is emitted as a photon.
You are correct that the photoelectric effect states that the kinetic energy of the ejected electron corresponds to the excess energy of the photon (energy beyond the work function), as has already been stated in three previous posts in this thread (see badmitondr, Econ, and mine).
But you are incorrect to say that if you try to excite an electron to a level using a photon of larger energy that it will emit the extra energy as a photon. This only occurs if the electron is
exactly excited to a higher excited state and then it falls back to a lower excited state. In such a case for instance, you might first absorb a 511 nm photon to go to an excited state II, which is short lived and falls down to excited state I, emitting a photon of let's say 916 nm (IR range). In that case there are two transition energies, each with an exact photon of the corresponding energy. For a good example of this, look at how a simply ruby laser works.