The power of a point $P$ wrt a circle $\omega$ is given by $\mathbb{P}(P, \omega) = PO^2 - r^2$ where $O$ and $r$ are the center and radius of $\omega$ respectively. The statement of linearity of power of a point is as follows:
If $\mathbb{P}(P, \omega_1, \omega_2) = \mathbb{P}(P, \omega_1) - \mathbb{P}(P, \omega_2)$ and $F : \mathbb{R}^2\mapsto\mathbb{R}$ so that $F := \mathbb{P}(P, \omega_1, \omega_2)$ then $F$ is linear.
My question is regarding whether or not the "converse" of this statement is true; that is, whether or not points satisfying the relation $F(C) = \frac{F(A) \cdot {CB} + F(B) \cdot {CA}}{{AB}}$ implies that the points $A, B, C$ in question are collinear.
No, the converse is not true. The affine space is $\mathbb{R}^2$, whereas the output of $F$ must be in $\mathbb{R}$. So, just because the output of $F(C)$ is a linear combination of $F(A)$ and $F(B)$ does not mean it must lie on the line. It is impossible for that to be true, since that would mean every value in $\mathbb{R}^2$ would have to have a unique value in $\mathbb{R}$.
Or another way of viewing it is to consider the barycentric definition of linearity of power of a point, where point $X$ with barycentric coordinates $X = (p,q,r)$ with respect to triangle $\triangle ABC$ satisfies $F(X) = pF(A) + qF(B) + rF(C)$. From this it’s clear that there are infinitely many points with some value of $F$, and most of them will not line on the line that you want.