We know that the set $D=\{a+b\sqrt{2} \mid a,b\in \mathbb Z\}$ is dense in $\mathbb R$ because $D$ is a subgroup of $(\mathbb R,+)$ that is not of the form $\alpha \mathbb Z$.
So, the following set $$\left\{(a+b\sqrt{2},c+d\sqrt{2}) \mid a,b,c,d\in \mathbb Z \right\}$$ is also dense in $\mathbb R^2$.
Here is my question : is it also true that the subset $$\left\{(a+b\sqrt{2},c+d\sqrt{2}) \mid a,b,c,d\in \mathbb Z, ad-bc=\pm 1 \right\}$$ is dense in $\mathbb R^2$ ?
If the answer is no, then can one find a vector $X\in \mathbb R^2$ such that $$\{MX \mid M\in GL_2(\mathbb Z)\}=\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}X \mid a,b,c,d\in \mathbb Z, ad-bc=\pm 1\right\}$$ is dense in $\mathbb R^2$ ?
It looks like the answer is positive. You can get a dense orbit from the subgroup $SL_2(\mathbb{Z})$ acting on any vector $v \in \mathbb{R}^2$ that isn't a scalar multiple of a vector in $\mathbb{Z} \times \mathbb{Z}$. See an elementary proof by S.G. Dani and Arnaldo Nogueira here. I'm not sure if there are proofs from scratch that are more brief.