$AB = I$, B is equal to the inverse of $A$ and vice versa.
So, $B= A^{-1}$
Now $CD =I$, $C$ is equal to the inverse of $D$ and vice versa.
So, $D =C^{-1}$
$=\left(A.\begin{bmatrix} 1& 0 \\ 1&1 \end{bmatrix}\right)^{-1}$
Remark: $(AB)^{-1} = B^{-1}A^{-1} $
$=\begin{bmatrix} 1& 0 \\ 1&1 \end{bmatrix}^{-1}.A^{-1}$
$=\begin{bmatrix} 1& 0 \\ {-1}&1 \end{bmatrix}.B$
$=\begin{bmatrix} b_{11}& b_{12} \\ b_{21}-b_{11}&b_{22}-b_{12} \end{bmatrix}$