(Not an answer.)
Let $r = \frac{1}{a}, s = \frac{1}{b}$. Reindex your points to $(x_0, y_0), ... (x_{n-1}, y_{n-1})$ and let $z_k^2 = x_k^2 + y_k^2$. Then $x_k = z_k \cos \frac{2\pi k}{n}, y_k = z_k \sin \frac{2\pi k}{n}$, and the intersection condition becomes
$$z_k^2 \left( r^2 \cos^2 \frac{2\pi k}{n} + s^2 \sin^2 \frac{2\pi k}{n} \right) = 1.$$
Together with the condition that $\prod_{k=0}^{n-1} z_k = 1$, it follows that the desired conditions can be stated as
$$\prod_{k=0}^{n-1} \left( r^2 \cos^2 \frac{2\pi k}{n} + s^2 \sin^2 \frac{2\pi k}{n} \right) = 1.$$
This is likely to be a hard Diophantine equation to solve in general. For $n = 3$, for example, the equation is
$$r^2 (r^2 + 3s^2)^2 = 16.$$
The curve $r(r^2 + 3s^2) = 4$ is an elliptic curve, and in general one must use computer algebra to rule out the existence of rational points on such curves. In this particular case we might be able to get away with some argument using unique factorization in $\mathbb{Z}[\omega]$, but this strategy will fail in general for the same reason it fails in Fermat's Last Theorem.