$ N:= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \ \ \sigma:= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\ \ \ $
$\Longrightarrow \ \ \ N^2=0, \ \ \ \sigma^2=1,\ \ \ \sigma N=N=-N\sigma$
So define
$A:=a\ N, \ \ \ C:=c\ N, \ \ \ D:=d\ N,$
$B:=b\ N+\beta\ \sigma$
$\Longrightarrow$ products among $\{A,C,D,...\}$ are zero, while $B^2=\beta^2\ 1$ and e.g. $BA=ab\ \sigma$.
The relation $\sigma N=N=-N\sigma$ makes $N$ an "eigenvector" of $\sigma$ from left and right so as to fulfill the anti-commutation relation $AB+BA=0$, without $AB$ already being zero. This way the expression involving $B$ is the only one which survives in your last condition:
$AD+BC=ad\ N^2+bc\ \sigma\propto \sigma\ne 0.$