Let the abscissae of the points $A$, $B$, $C$, $D$, $F$ and $G$ be $a$, $b$, $c$, $d$, $f$ and $g$ respectively. If $O_1(\alpha, \beta)$ is the center of the circle $S_1$, the coordinates of the points $A$, $B$, $C$ and $D$ satisfy the system of equations
$$\begin{cases} (x - \alpha)^2 + (y - \beta)^2 = R^2, \\ y = \frac{1}{x}. \end{cases}$$
Eliminating $y$ we obtain the equation
$$x^4 - 2\alpha x^3 + (\alpha^2 + \beta^2 - R^2)x^2 - 2\beta x + 1 = 0,$$
with solutions $a$, $b$, $c$ and $d$. Whence $a + b + c + d = 2\alpha$ and $abcd = 1$. Similarly, if $O_2(\lambda, \mu)$ is the center of the circle $S_2$, then $a + b + f + g = 2\lambda$ and $abfg = 1$. Since the radii of the circles $S_1$ and $S_2$ are equal, the quadrilateral $AO_1BO_2$ is a rhombus, hence the midpoints of $AB$ and $O_1O_2$ coincide, therefore $\frac{a + b}{2} = \frac{\alpha + \lambda}{2}$ which gives $a + b = \alpha + \lambda$. So
$$2\alpha + 2\lambda = (a + b + c + d) + (a + b + f + g) = 2\alpha + 2\lambda + c + d + f + g,$$
whence $f + g = -(c + d)$. Moreover, $cd = fg = (ab)^{-1}$. Rewrite it as $f + g = (-c) + (-d)$ and $f \cdot g = (-c) \cdot (-d)$. Thus the pairs $(f, g)$ and $(-c, -d)$ have equal sums and products. So either $f = -c$ and $g = -d$ or $f = -d$ and $g = -c$. If, for example, $f = -c$ and $g = -d$, then the pairs $F, C$ and $G, D$ of points are symmetric with respect to the point of origin, i.e. $FGCD$ is a parallelogram with the center at the origin.