The error is that you assume that the density distribution is "nearly spherically symmetric". It's far enough from spherical symmetry if you want to calculate first-order subleading effects such as the equatorial bulge. In other words, the term $hg$ in your potential is wrong.

Just imagine that the Earth is an ellipsoid with an equatorial bulge, it's not spinning, and there's no water on the surface. What would be the potential on the surface? You have de facto assumed that in this case, it would be $-GM/R+h(\theta)g$. However, by this Ansatz, you have only acknowledged the variable distance of the probe from a spherically symmetric source of gravity: you have still neglected the bulge's contribution to the non-sphericity of the gravitational field.

If you include the non-spherically-symmetric correction to the gravitational field of the Earth, $hg$ will approximately change to $hg-hg/2=hg/2$, and correspondingly, the required bulge $\Delta h$ will have to be doubled to compensate for the rotational potential. I will try to add more accurate formulae for the gravitational field of the ellipsoid in an updated version of this answer.