Denote by $k$ the incircle and by $k_{a}$ the excircle opposite to $A$ of triangle $ABC$. Let $k$ and $k_{a}$ touch the side $BC$ at the points $X$ and $T$, respectively, let $k_{a}$ touch the lines $AB$ and $AC$ at the points $P$ and $Q$, respectively. We use several times the fact that opposing sides of a parallelogram are of equal length, that points of contact of the excircle and incircle to a side of a triangle lie symmetric with respect to the midpoint of this side and that segments on two tangents to a circle defined by the points of contact and their point of intersection have the same length. So we conclude
$$\begin{gathered} ZP = ZB + BP = XB + BT = BX + CX = ZS \text{ and } \\ CQ = CT = BX = BZ = CS. \end{gathered}$$

So for each of the points $Z, C$, their distances to $S$ equal the length of a tangent segment from this point to $k_{a}$. It is well-known, that all points with this property lie on the line $ZC$, which is the radical axis of $S$ and $k_{a}$. Similar arguments yield that $BY$ is the radical axis of $R$ and $k_{a}$. So the point of intersection of $ZC$ and $BY$, which is $G$ by definition, is the radical center of $R, S$ and $k_{a}$, from which the claim $GR = GS$ follows immediately.

Denote $x = AZ = AY$, $y = BZ = BX$, $z = CX = CY$, $p = ZG$, $q = GC$. Several lengthy calculations (Menelaos' theorem in triangle $AZC$, law of Cosines in triangles $ABC$ and $AZC$ and Stewart's theorem in triangle $ZCS$) give four equations for $p, q, \cos \alpha$ and $GS$ in terms of $x, y$, and $z$ that can be resolved for $GS$. The result is symmetric in $y$ and $z$, so $GR = GS$. More in detail this means:
The line $BY$ intersects the sides of triangle $AZC$, so Menelaos' theorem yields $\frac{p}{q} \cdot \frac{z}{x} \cdot \frac{x+y}{y} = 1$, hence
$$\begin{equation*} \frac{p}{q} = \frac{x y}{y z + z x} . \tag{1} \end{equation*}$$
Since we only want to show that the term for $GS$ is symmetric in $y$ and $z$, we abbreviate terms that are symmetric in $y$ and $z$ by capital letters, starting with $N = x y + y z + z x$. So (1) implies
$$\begin{equation*} \frac{p}{p+q} = \frac{x y}{x y + y z + z x} = \frac{x y}{N} \quad \text{ and } \quad \frac{q}{p+q} = \frac{y z + z x}{x y + y z + z x} = \frac{y z + z x}{N} . \tag{2} \end{equation*}$$
Now the law of Cosines in triangle $ABC$ yields
$$\cos \alpha = \frac{(x+y)^2 + (x+z)^2 - (y+z)^2}{2(x+y)(x+z)} = \frac{2x^2 + 2x y + 2x z - 2y z}{2(x+y)(x+z)} = 1 - \frac{2 y z}{(x+y)(x+z)} .$$
We use this result to apply the law of Cosines in triangle $AZC$:
$$\begin{align*} (p+q)^2 & = x^2 + (x+z)^2 - 2x(x+z) \cos \alpha \\ & = x^2 + (x+z)^2 - 2x(x+z) \cdot \left(1 - \frac{2 y z}{(x+y)(x+z)}\right) \\ & = z^2 + \frac{4 x y z}{x+y} \tag{3} \end{align*}$$
Now in triangle $ZCS$ the segment $GS$ is a cevian, so with Stewart's theorem we have $p y^2 + q(y+z)^2 = (p+q)\left(GS^2 + p q\right)$, hence
$$GS^2 = \frac{p}{p+q} \cdot y^2 + \frac{q}{p+q} \cdot (y+z)^2 - \frac{p}{p+q} \cdot \frac{q}{p+q} \cdot (p+q)^2 .$$
Replacing the $p$'s and $q$'s herein by (2) and (3) yields
$$\begin{aligned} GS^2 & = \frac{x y}{N} y^2 + \frac{y z + z x}{N}(y+z)^2 - \frac{x y}{N} \cdot \frac{y z + z x}{N} \cdot \left(z^2 + \frac{4 x y z}{x+y}\right) \\ & = \frac{x y^3}{N} + \underbrace{\frac{y z (y+z)^2}{N}}_{M_1} + \frac{z x (y+z)^2}{N} - \frac{x y z^3 (x+y)}{N^2} - \underbrace{\frac{4 x^2 y^2 z^2}{N^2}}_{M_2} \\ & = \frac{x y^3 + z x (y+z)^2}{N} - \frac{x y z^3 (x+y)}{N^2} + M_1 - M_2 \\ & = \underbrace{\frac{x\left(y^3 + y^2 z + y z^2 + z^3\right)}{N}}_{M_3} + \frac{x y z^2 N}{N^2} - \frac{x y z^3 (x+y)}{N^2} + M_1 - M_2 \\ & = \frac{x^2 y^2 z^2 + x y^2 z^3 + x^2 y z^3 - x^2 y z^3 - x y^2 z^3}{N^2} + M_1 - M_2 + M_3 \\ & = \frac{x^2 y^2 z^2}{N^2} + M_1 - M_2 + M_3, \end{aligned}$$
a term that is symmetric in $y$ and $z$, indeed.