$M= \{ A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 \}$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$

The group $M$, which I am going to rename $G$, acts transitively on $\mathbb{R}^2 \setminus \{(0, 0)\}$ which is clearly homeomorphic to $\mathbb{R} \times S^1$. The stabilizer of the vector $$ \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$ is the set of matrices of the form $$ \begin{pmatrix} 1 & * \\ 0 & 1 \end{pmatrix} $$ which is clearly homeomorphic to $\mathbb{R}$.

If you've seen group actions, that's a complete answer.

EDIT: I translated this into a language not using group actions in my previous answer, but made quite a mistake (the equations I wrote always admit a trivial solution).