What is the mechanism of endothermic dissolution?
Since the temperature of the solid is now slightly lower than the temperature of the surrounding water, heat will be transferred from the water to the solid until the temperatures are equal

It is fine to think that way as a rough draft in rationalizing this process. The only issue is that temperature is a macroscopic quantity, and mechanism tends to be a particular (molecular) concept. So you could say that the individual solid particles (e.g. ions) will have slightly lower kinetic energy than average, and collisions with individual solvent molecules will balance this out.
Or you could start with solvent molecules near the solid-liquid interface that already have a higher kinetic energy than the average become part of the solid, thus lowering the average kinetic energy of the remaining solute molecules.
Or you could say that the water molecules solvating the ions use parts of their kinetic energy in the desolvation process, directly lowering the temperature of the solvent.
Because all the species have a distribution of kinetic energies around the average, all these scenarios are correct.

As a result, the temperature of the remaining solid will drop a little.

I don't think the bulk of the solid will have a lower average kinetic energy, so I don't think the temperature of the solid will drop faster than that of the solution. I would think that there is a temperature gradient between the solid-liquid interface on the one hand, and the bulk solid and liquid on the other hand. This is maintained as long as there is a net transfer from solution to solid. Once equilibrium is reached, of course, this will include thermal equilibrium.

this energy will be sufficient to overcome the 4 kJ/mol barrier

&quot;Barrier&quot; sounds like an activation energy, something you can step over. Here, I would rather talk of &quot;energetic cost&quot; or &quot;step up&quot;, something that you don't get back. The solvation energy is higher than the lattice energy, so there is a cost for taking the ions out of solution and adding them to the ionic solid.