You can analyze this either in the rotating frame or in an inertial frame.
The analysis in the rotating frame is simple. In this frame, nothing is moving, so the surface of the earth, which is in hydrodynamic equilibrium, is an equipotential. The gravitational time dilation depends on the difference in potential, so it's zero.
You're doing the analysis in the inertial frame and trying to analyze the structure of the earth explicitly, which makes it quite a bit more complicated. In this frame, it is not true that the earth's surface is an equipotential. (Not only that, but it is not true that when you take a small mass from the pole to the equator, the sum of PE+KE for that mass is conserved -- in fact, both increase. This is balanced only by the reduction in KE of the rest of the earth, whose rotation slows down by conservation of angular momentum.)
The expression you're starting with for the equatorial-polar flattering is actually wrong for a self-gravitating, incompressible fluid. There is a nice treatment of the topic by Fitzpatrick here: https://farside.ph.utexas.edu/teaching/336k/Newton/node109.html One thing to watch out for is that you can't find the difference in potential between the pole and equator by using the potential of a sphere. In fact the potential of a uniform ellipse is different from that of a sphere, and the effect is of leading order. This produces a factor of 5/4, which is missing from your analysis.
On top of this is the fact that the earth's density is nonuniform. Therefore the polar flattening comes out to be different from the incompressible fluid result by about a factor of 1.28.
These two errors in your calculation almost cancel out, but they result in a net error of 2% in your calculation relative to what is actually observed for the earth.
So then you take this result for the flattening, which is fortuitously approximately right, and plug it in to the equation for the potential of a uniform sphere. But the potential isn't actually that of a uniform sphere, so you end up with a result that is wrong.
So although it is exactly true that the time dilation cancels, it's somewhat complicated to prove that by explicit calculation in the inertial frame, for a self-gravitating, rotating, incompressible fluid -- and the complication would be even greater if you wanted to model the actual earth, which is nonuniform.
If you want to see a little more explicitly that the cancellation of the relativistic effect always works to leading order and is independent of assumptions like incompressibility, a nice way to do it is the following. In Fitzpatrick's notation, let $\Phi$ be the Newtonian potential in the inertial frame. The field is minus the gradient of this. When we switch to the rotating frame, the potential changes by a term $\chi=-(1/2)\Omega^2 r^2\sin^2\theta=-(1/2)v^2$, whose gradient is the centrifugal acceleration. In the rotating frame, hydrodynamic equilibrium requires that $\Phi+\chi$ be constant for the surface of the earth. But this is equivalent to $\Phi-(1/2)v^2$ being constant, so that the gravitational and kinematic time dilations cancel to leading order, and this cancellation is exact in the sense of being independent of the detailed structure of the earth.