Points A and B have a common Y coordinate. For each point P on the curve, there is a pair of points A and B such that A-B=P. Does this give us an advantage when computing the private key?
Yes. But just when the only information you have is that the private key is anywhere between 1 and n - 1.
In other words: good luck finding the specific pair of points that add up to P (or one of its equivalent X endos).