The sky does not skip over the green range of frequencies. The sky is green. Remove the scattered light from the Sun and the Moon and even the starlight, if you so wish, and you'll be left with something called airglow (check out the link, it's awesome, great pics, and nice explanation).
Because the link does such a good job explaining airglow, I'll skip the nitty gritty.
So you might be thinking, "Jim, you half-insane ceiling fan, everybody knows that the night sky is black!" Well, you're only half right. The night sky isn't black. The link above explains the science of it, but if that's not good enough, try to remember back to a time when you might have been out in the countryside. No bright city lights, just the night sky and trees. Now when you look at the horizon, can you see the trees? Yes, they're black silhouettes against the night sky. But how could you see black against black? The night sky isn't black. It's green thanks to airglow (or, if you're near a city, orange thanks to light pollution).
Stop, it's picture time. Here's an above the atmosphere view of the night sky from Wikipedia:
And one from the link I posted, just in case you didn't check it out:
See, don't be worried about green. The sky gets around to being green all the time.
The answer to your question is the obverse of it: we assign a color to an object based on the wavelengths which are reflected to our eyes (or in the case of filters, transmited to our eyes). That means other wavelengths are absorbed. The absorption of wavelengths is based, primarily, on the chemistry of the object.
Red dye applied to cotton cloth is a chemical whose molecules absorb less red light than other wavelengths, hence the red wavelengths are more intense than other wavelengths in comparison to the light from other objects. Similarly for blue, green, yellow, etc objects. Most objects of colors don't absorb all the energy of other wavelengths; they just absorb less of certain wavelengths, and we assign a color name based on the modified mixture reaching our eyes.
In fact, the "colors" surrounding each other can modify our interpretation of what color we see. (Search for "color optical illusions". There are fascinating examples.)
Regarding absorb and reflect: they mean exactly what you think. The energy of an EM wave is taken into a molecular structure and not released as the same wavelength (absorption) or it is released as the same wavelength (reflection or transmission).
Best Answer
Reflection can be calculated in a completely classical way requiring only that the material have a bulk polarisibility. It's true that ultimately the polarisibility is a result of the electron configuration, and this arises from quantum mechanics, but you would not call reflection a quantum process any more than you'd call viscosity of a fluid a quantum process.
By contrast, fluorescence is exclusively quantum because it requires electrons to be excited into a higher energy state, then lose some of their energy in lattice interactions and settle to a lower level state. The emission is then from electrons dropping from this slightly lower energy state to the ground state. This is why the fluorescence wavelength is always lower than the wavelength of the incident light.
A quick note on terminology, you say:
but fluorescence is inelastic scattering, while reflection is elastic scattering (specifically coherent elastic scattering). Fluorescence is inelastic because light is absorbed in an electronic trasnition and some of the energy lost to the lattice. Reflection is elastic because it does not involve an electronic transition and no energy is lost.