Yes, you can formalize this intuition. You can look for solutions over the algebraic closure of the field $\mathbb Q(x_0)$, setting $y=x_0$. If there are just $k$ solutions, then at every point except for a positive codimension closed set there will just be $k$ solutions of the original problem.
The reason this works is that, if you consider the projection from the variety of points satisfing $Q(y)P(x)=Q(x)P(y)$, $x,y\not\in V$ to the variety of points $x$ satisfying $x\not \in P$V$, the fiber over a typical point (meaning, a point not on a specific closed subset of positive codimension) will be geometrically the same as the fiber over the generic point. One geometric property of a variety in this sense is the number of points it has over the algebraic closure of the base field.
If the fiber over the generic point has extra points, but for some reason these usually fail to be real or fail to lie in $[0,1]^n$, proving that is more subtle, and ideas of genericity may not be as helpful.
This is all assuming you are looking for real solutions. Things are different with rational solutions.
