No, that is not true. For instance, let $k$ be $\mathbb{R}$, let $X$ be $\text{Spec}(\mathbb{R})$, let $Y$ be $\text{Spec}(\mathbb{C})$, and let $G$ be the multiplicative group. Compare the real points of $R_{Y/X}(G_Y)$ with its induced analytic topology to the real points of $\mathbb{G}^2_X$. The homotopy groups $\pi_0$ and $\pi_1$ distinguish these two real Lie groups immediately.

No, that is not true. For instance, let $k$ be $\mathbb{R}$, let $X$ be $\text{Spec}(\mathbb{R})$, let $Y$ be $\text{Spec}(\mathbb{C})$, and let $G$ be the multiplicative group, $\mathbb{G}_m$. Compare the real points of $R_{Y/X}(G_Y)$ with its induced analytic topology to the real points of $\mathbb{G}^2_{m,X}$. The homotopy groups $\pi_0$ and $\pi_1$ distinguish these two real Lie groups immediately.