Here is a solution that you can find by solving a linear system. First we will obtain a $4\times 4$ matrix that looks like your target matrix ($I_2$) but padded. We let
$A = \begin{bmatrix}a_1 & 0 & 0 & 0\\0 & a_2 & 0 & 0\\0 & 0 & a_3 & 0\\0 & 0 & 0 & a_4\end{bmatrix} \quad{}\text{and}\quad X = \begin{bmatrix} x_{1,1} & x_{1,2} & x_{1,3} & x_{1,4} \\ x_{2,1} & x_{2,2} & x_{2,3} & x_{2,4} \\ x_{3,1} & x_{3,2} & x_{3,3} & x_{3,4} \\ x_{4,1} & x_{4,2} & x_{4,3} & x_{4,4} \end{bmatrix}$
When you multiply these two two matrices you obtain
$AX = \begin{bmatrix} a_1 x_{1,1} & a_1 x_{1,2} & a_1 x_{1,3} & a_1 x_{1,4} \\ a_2 x_{2,1} & a_2 x_{2,2} & a_2 x_{2,3} & a_2 x_{2,4} \\ a_3 x_{3,1} & a_3 x_{3,2} & a_3 x_{3,3} & a_3 x_{3,4} \\ a_4 x_{4,1} & a_4 x_{4,2} & a_4 x_{4,3} & a_4 x_{4,4} \end{bmatrix} $
You want all the entries of this matrix to be $0$ except the four entries in the upper left corner (which we want to be $a_1,a_2,a_3$ and $a_4$). Solving the equations formed leads to
$ X = \begin{bmatrix} 1 & a_2/a_1 & 0 & 0 \\ a_3/a_2 & a_4/a_2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $ and when you multiply $A$ and $X$ together you obtain
$AX = \begin{bmatrix} a_1 & a_2 & 0 & 0 \\ a_3 & a_4 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
Note that the solution assumes that $a_1 \neq 0$ and $a_2 \neq 0$. If one or more of these is zero, we cannot solve the system. In your case it is okay because you say that the entries on the diagonal are non-zero. As a final step we need to remove two of the columns and two of the rows. Let $B$ be given by
$B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}.$
Then $B^T A X B = I_2$. Note that the matrix $X$ depends on the entries of matrix $A$. I am not sure if there is a matrix that will work on all such diagonal matrices.