I think the answer to your question is negative. Consider for instance
$$M = \begin{pmatrix}
1 & 0 \\ 
0 & 1
\end{pmatrix}, \quad N = \begin{pmatrix}
0 & 1 \\ 
1 & 0
\end{pmatrix}$$
Assume that there exist 
$$A = \begin{pmatrix}
a_{11} & a_{12} \\ 
0 & a_{22}
\end{pmatrix}, \quad B = \begin{pmatrix}
b_{11} & b_{12} \\ 
0 & b_{22}
\end{pmatrix}$$
with $\det(A) = \det(B) = 1$ and such that
$$A\cdot M\cdot B^{T} = N$$
Then
$$A\cdot B^{T} = \begin{pmatrix}
a_{11}b_{11}+a_{12}b_{12} & a_{12}b_{22}\\ 
a_{22}b_{12} & a_{22}b_{22}
\end{pmatrix} = \begin{pmatrix}
0 & 1\\ 
1 & 0
\end{pmatrix}$$
and hence either $a_{22} = 0$ or $b_{22}=0$ which contradict $\det(A) = \det(B) = 1$. 

More generally, your action stabilizes the locus of matrices of the form $\begin{pmatrix}
m_{11} & m_{12}\\ 
m_{21} & 0
\end{pmatrix}$.