Suppose $A\in \mathrm{M}_2(\mathbb{Z})$ has determinant $r\neq 0$. I want to show there exists matrices $M,N \in \mathrm{SL}_2(\mathbb{Z})$ such that $$A = M \begin{bmatrix}r & 0 \\ 0 & 1 \end{bmatrix} N$$
Since $\mathrm{SL}_2(\mathbb{Z})$ is generated by $S= \begin{bmatrix} 0 & -1 \\ 1 &0 \end{bmatrix}$ and $T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, we can find $M_1\in\mathrm{SL}_2(\mathbb{Z})$ such that $M_1A$ is upper triangular. Then you can conjugate by $S$ to get a lower triangular matrix. Repeat the above process to get matrices $M_2,N_2$ such that $M_2AN_2$ is diagonal. But I don't know how to finish it off. Any help would be appreciated.
EDIT: As pointed out in the comments. Unless $A$ is primitive (gcd of the entries is 1), the statement is false since the Smith Normal Form is unique. If we suppose that $A$ is primitive, can I finish the above argument?