How are solubilities of hydrate salts determined when competing equilibria are unavoidable?
Let there be three species: a hydrated solid $H$, an anhydrous solid $A$, and a complete solvated (liquid) aqueous solution, $S$.

Your question is about differences in the two equilibria $\ce{H &lt;=&gt; S}$ and $\ce{A &lt;=&gt; S}$.  Can $H$ really have a different solubility than $A$?

I think the answer is yes, but only if the equilibrium $\ce{H&lt;=&gt;A}$ cannot and is not established.  In the presence of an "infinite" kinetic barrier that blocks conversion of the hydrate to the anhydrous form and vice versa, it makes sense to treat $\ce{H&lt;=&gt;S}$ and $\ce{H&lt;=&gt;A}$ and independent equilibria which might have different equilibrium constants.

In a comment to your question, I made it seem like "any" precipitation from a (super)saturated solution of $S$ would be the hydrate.  On further reflection, I think that is true, unless (i) the equilibrium $\ce{A&lt;=&gt;H}$ cannot be established due to kinetic barriers, and (ii) there is solid phase $A$ present in the solution.  In that case, the supersaturation could just result in "precipitation" of solid onto the existing crystal surface of already-present $A$, a process that would increase the mass of $A$ (solid) in the system but which would not require formation of any crystallites or nucleation intermediates.


  How, then, was the solubility of the hydrate ever experimentally determined?


I think the answer is that the hydrate solubility can differ from the anhydrous solid solubility only in cases when there is no mechanism by which equilibrium can be established between the hydrate and the anhydrous solid.