I'm not sure I quite understand what the question is. But you certainly don't sound stupid for asking it.
A photoexcited electron does return to its ground state, but unless you're talking about isolated atoms/ions, it usually does so via a different pathway.
When an electron in a molecular orbital absorbs a photon of light (of some energy in the UV or visible) and is excited to another electronic state, there is a natural driving force to return to a state of low potential energy. One of the ways this can happen is for the electron to re-emit a photon or nearly the same energy that took to excite it in the first place. This phenomenon we call fluorescence.
However most molecules do not manifest observable fluorescence. This is because there is usually coupling between the excited electronic state and the ground electronic state mediated by molecular vibrational wavefunctions, and the rate of fluorescence is typically slower than the rate of conversion of electronic excitation to vibrational excitation, a process called internal conversion. Essentially what happens is that instead of emitting a photon of light, the molecule disperses all this excess energy through heat loss to the surrounding environment. Internal conversion is almost always fast compared to fluorescence - except under the condition of high rigidity and under the condition of very large energy gaps between the ground and excited electronic states. This is why there are far more blue-fluorophores than red fluorophores, and it is why most fluorophores tend to be large flat molecules without a lot of "floppy" functional groups. (Amino groups, for example, are notorious for quenching fluorescence.) The specific reasons that rigidity and high energy gap play a role in the fluorescence (quantum) yield are fairly complicated but I would be happy to elaborate if you are interested. Do note that no matter whether the molecule is highly fluorescent or not, there is always an interplay between fluorescence and internal conversion - the relative rates of these two processes determine the overall fluorescence yield (number of photons emitted vs. number of photons absorbed). Like most things it's a matter of probability.
(Also note that fluorescence and internal conversion aren't the only processes that can deactivate the excited electronic state. Photochemistry can, as can intersystem crossing to an excited triplet surface; triplet states can also relax to the ground state via internal conversion or by emission of a light photon. The latter process is called phosphorescence and due to the forbidden nature of this transition it is typically prolonged and can linger seconds or minutes after the light sources is removed. This is how glow-in-the-dark stickers work. Ironically phosphorous, after which the process is named, does not emit light via phosphorescence.)
Having said all that, even in a molecule with a quantum yield of 100% (every photon absorbed is emitted as a photon) - and there are some, like rhodamine, that can come pretty close to this - the light emitted never is the same color as the light absorbed. For one thing, absorption can result in population of many electronic states, but typically fluorescence only occurs from one of them. This is called Kasha's Rule.http://en.wikipedia.org/wiki/Kasha%27s_rule
Although in grad school we cheekily liked to call it Kasha's Good Idea, because it's not always obeyed. This is why an absorption spectrum can have many peaks but a fluorescence spectrum typically only has one. So for instance if you irradiate a molecule with blue light of 400 nm, you may actually be populating a high-lying singlet state rather than the lowest-lying excited electronic singlet. That state will relax to the lowest lying excited singlet within a few femtoseconds and from there you get fluorescence of red light.
Even if you specifically excite into the lowest energy absorption band, though, the emission wavelength is still red-shifted due to a Stokes shift. This may only be a few nanometers or can be very large, depending on the molecule and the solvent environment. This happens because when you excite an electron in a molecule, all of the electron density in the molecule changes. Usually this results in polarization of the electron cloud, which can result in very different polarity of molecules when they are in their excited states versus their ground states. Light absorption is nearly instantaneous compared to nuclear motion (embodied in the Franck-Condon principle, which also explains why vibronic transitions vary in intensity), but if the excited state lives long enough, nearby solvent molecules will reorganize themselves around the polarized, excited molecule, which will in turn stabilize the excited state. This results in a shrinking of the energy gap, which manifests itself is a lower-energy emission wavelength than the absorption wavelength. In addition, structural relaxation of the excited molecule itself can occur, which can lower the excited-state energy even more.
So, to sum. Excited state relaxation dynamics are complicated but almost always result in a lowering of the energy of the excited-state prior to emission, so the fluorescence color is almost always different from the absorption color. That's if there is fluorescence at all, which most times there isn't. The exception to this rule about red-shifting is atomic fluorescence, which have practically zero Stokes shift because there is no structural relaxation in an atom. Some laser dyes and other highly rigid molecules have very small Stokes shifts (and high quantum yields), but even in these cases you can see a color difference with even the most basic spectrophotometric equipment. And even in the idealized case where the emission color was EXACTLY the same color at the absorption color, fluorescence is in a random direction. So if you were looking at a sample that did emit 100% of its absorbed photons at EXACTLY the same energy, the substance would STILL appear colored to your eye because you'd only detect a small fraction of the emitted photon with your eye.
Well that was a shotgun approach to your question. Maybe I hit it with something in there. If not, please clarify the question and I'll try to narrow the answer.