One can see that $I=(f_1,\dots, f_s)$ is $G$-invariant iff $\alpha\cdot f_i \in I$ for each $\alpha \in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and $I+J$. Thus, for example, $(x,y)^n$ are invariant for all $n\geq 0$. One also have that $I=(x^{2^k}, y^{2^k})$ is invariant as pointed out by YCor. So you can generate many other examples. A complete classification seems difficult though.

One can see that $I=(f_1,\dots, f_s)$ is $G$-invariant iff $\alpha\cdot f_i \in I$ for each $\alpha \in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and $I+J$. Thus, for example, $(x,y)^n$ are invariant for all $n\geq 0$. One also have that $I=(x^{2^k}, y^{2^k})$ is invariant as pointed out by YCor. So you can generate many other examples. A complete classification seems difficult though.

For the case of two-generated ideals, one could use the above remark and the fact that "Frobenius commutes with linear change of variables" to show:

1. If $I=(f,g)$ is invariant then $J= (f^{2^k}, g^{2^k})$ is invariant.

2. If $I=(f,g)$ is invariant and $\deg(f)> \deg(g)$ then $J=(f^{2^k}, g^l)$ is invariant with $l\leq 2^k$.

For instance the ideal $(x^6, x^2+xy+y^2)$ is invariant.