One can prove the identity with induction on the dimension $n$, using elementary transformations on the determinant.
The identity holds for $n=0$, with the empty product and the empty determinant both being equal to $1$.
For the induction step $n \to n+2$ we compute
$$
R_{n+2}(x_1, \ldots, x_n, a, b) =
\begin{vmatrix}
q(x_1, x_1) & \cdots & q(x_1, x_n) & q(x_1, a) & q(x_1, b) \\
\vdots & \ddots & \vdots &\vdots &\vdots \\
q(x_n, x_1) & \cdots & q(x_n, x_n) & q(x_n, a) & q(x_n, b) \\
q(a, x_1) & \cdots & q(a, x_n) & 0 & q(a, b) \\
q(b, x_1) & \cdots & q(b, x_n) & q(b, a) & 0
\end{vmatrix}
$$
where I have used
$$
q(x, y) = \frac{x-y}{x+y}
$$
to shorten the expressions a bit. We add multiples of the last two rows to the first $n$ rows to eliminate the last two entries in those rows. This gives
$$
R_{n+2}(x_1, \ldots, x_n, a, b) =
\begin{vmatrix}
d_{1,1} & \cdots & d_{1,n} & 0 & 0 \\
\vdots & \ddots & \vdots &\vdots &\vdots \\
d_{n,1} & \cdots & d_{n,n} & 0 & 0 \\
* & \cdots & * & 0 & q(a, b) \\
* & \cdots & * & q(b, a) & 0
\end{vmatrix}
$$
with
$$
d_{i,j} = q(x_i, x_j) - \frac{q(x_i, a)q(b,x_j)}{q(b, a)} - \frac{q(x_i, b)q(a,x_j)}{q(a, b)} \, .
$$
Now the magic (?) happens: This expression simplifies to
$$
d_{i,j} = q(x_i, a)q(x_i, b)q(x_j, a)q(x_j, b)q(x_i, x_j) \, .
$$
The common factors $q(x_i, a)$ and $q(x_i, b)$ can be extracted from each row and each column in $ \det \{ d_{i,j} \}$, and it follows that
$$
R_{n+2}(x_1, \ldots, x_n, a, b) = \det \{ d_{i,j} \} q(a, b)^2 \\
= \det \{ q(x_i, x_j) \}\left(\prod_{i=1}^n q(x_i, a)^2 q(x_i, b)^2\right) q(a, b)^2 \\
= R_{n}(x_1, \ldots, x_n)\left(\prod_{i=1}^n q(x_i, a)^2 q(x_i, b)^2\right) q(a, b)^2 \, ,
$$
which is exactly the induction step.
 
instead of the (ab)using the$\,$
TeX command, that looks a bit nicer in a link. $\endgroup$