Some additional information relevant to this question.
I'm going to use "forces" and "particles" in this explanation to generalize it. Particles can be ions (ionic compounds held together by strong electrostatic forces), metal atoms (held together by electrostatic forces from attraction of the metal atom nuclei for the "sea of electrons") or molecules (for covalent compounds held together in the solid or liquid by intermolecular forces: London dispersion forces, dipole-dipole attractions, and hydrogen bonding. Note: we are not talking about the intramolecular covalent bonds in this explanation.) So, solids and liquids are held together in that form by forces between the particles.
The reason that melting is always endothermic (except for the exceptions given by F'x...which I also did not know about) is that when you melt a substance, you have to break up some of the forces holding the particles together in the solid so that they can move past each other and form a liquid. This takes energy, the "heat of fusion" (usually given on a per mole basis) mentioned above. The stronger the forces, the more energy it would take to melt the substance. As long as you are melting a substance and holding the temperature at the melting point (i.e. not adding enough energy to raise the temperature) then you will have an equilibrium between the solid and liquid.
I think there is some confusion in the terms you are using. Desolvation means removing the solvent from a solute-solvent mixture. (This can happen in crystallization, for example.) Solvation (which I think you're asking about) is the interaction of solvent particles with solute particles. Dissolving or dissolution is the process in which a solid, liquid, or gas forms a solution with a solvent.
So, as answered by Ashu, dissolution and be either exothermic or endothermic as he's noted above. This is true because in order to dissolve a substance in a solute, it takes energy to break up the solvent-solvent interactions, it takes energy to break up the solute-solute interactions, and you get energy back from new solvent-solute interactions. If you have stronger attractions between the solvent and solute than you had originally, then the process releases energy and is exothermic. The same concepts can be applied to mixing two liquids...which is usually referred to by saying they are "miscible" instead of speaking of one dissolving in the other. Ethanol and water are miscible in all proportions.
Can alcohol dissolve ice ("at or just below the melting point of ice" as phrased in your original question)? There are a couple of ways to answer this. If you put ice in ethanol at 0 degrees (its melting point) it will melt because the melting point of ice in a mixture of water and ethanol is lower than the melting point of pure water. The melting point depends on the proportions of water and ethanol.(That is to say, as some melted, it could not refreeze as you would expect to happen in an equilibrium mixture, so it would continue to melt.) See colligative properties for more information.
You've given two equations, H2O(s)→H2O(l)
H2O(s)+nEtOH(l)→H2O⋅EtOHn
In either case, you will need to provide the heat of fusion to melt the ice. So the question you asked is really about the energy of solvation of water in ethanol. When you mix water and ethanol, the flask feels warmer, so this is an exothermic process; you get more energy back than you put into it because of good interactions (hydrogen bonding) between water and ethanol particles. That solvation energy would provide some of the energy needed to melt the ice.
For the broader question "can something happen at a given temperature", you're generally asking if it happens spontaneously. To answer this you'd have to calculate the Gibbs Free Energy. This would take into account not only the energy considerations spoken of above, but also entropy considerations (which would matter here). If the Gibbs Free Energy value is negative, then the process or reaction is spontaneous at the temperature given. So, the answer to which would be the preferred process would also depend on entropy considerations.