Combining LSpice and my comments into an answer: If $G(k)$ is Zariski dense in $G$, then $A^{G(k)} = A^G$. It is very common that $G(k)$ is Zariski dense in $G$: This happens whenever (1) $k$ is infinite (2) $G$ is connected and either (3a) $G$ is reductive or else (3b) $k$ is perfect. See here.

Combining LSpice's (1 2) and my (1 2) comments into an answer: If $G(k)$ is Zariski dense in $G$, then $A^{G(k)} = A^G$. It is very common that $G(k)$ is Zariski dense in $G$: This happens whenever (1) $k$ is infinite and (2) $G$ is connected and either (3a) $G$ is reductive or else (3b) $k$ is perfect. See Density question in algebraic group.