Using the notation from that MSE answer, the task reduces to finding the locus $q\in S^3$ such that, for fixed $a,b\in S^3$,
$$\arccos(2\langle q,a\rangle^2-1)=\arccos(2\langle q,b\rangle^2-1)\text{.}$$
Since $\arccos$ is a bijection between $[-1,1]$ and $[0,\pi]$, the equality holds iff
$\langle q,a\rangle^2-\langle q,b\rangle^2=0$. This is last equality involves a difference of squares; consequently, equality holds iff
$$\begin{align}\langle q,a+b\rangle&=0 &&\text{or} &\langle q,a-b\rangle&=0\end{align}\text{.}$$
Now, $\langle q,c\rangle=0$ is the equation of a hyperplane passing through the origin. The intersection of such a hyperplane with $S^3$ is a great sphere $S^2$, so in $S^3$ the solution locus is $S^{2}\coprod_{S^1}S^{2}$, two great spheres that intersect along the great circle determined by the intersection of the plane $\langle q,a\rangle=\langle q,b\rangle=0$ with $S^3$.

Now we use that $S^3$ is a double cover of $\mathbf{RP}^3\cong\mathrm{SO}(3)$ by identifying antipodal points. Thus, the solution locus in $\mathrm{SO}(3)$ is $\mathbf{RP}^{2}\coprod_{\mathbf{RP}^1}\mathbf{RP}^{2}$—note that the two great spheres are *not* identified.