I've made a stupid mistake somewhere, but I can't find it. Maybe I'm misunderstanding something.
Work so far:
Let A, B $\in \mathbb{R^*}$
$$A= \begin{pmatrix} r & 0 & 0 \\ 0 & 1 & 0 \\ r-s & 0 & s \\ \end{pmatrix} $$
and
$$B = \begin{pmatrix} t & 0 & 0 \\ 0 & 1 & 0 \\ t-u & 0 & t \\ \end{pmatrix} $$
$$AB = \begin{pmatrix} rt & 0 & 0 \\ 0 & 1 & 0 \\ rt-su & 0 & st \\ \end{pmatrix} $$
The matrix is of the form: $$\begin{pmatrix} a & 0 & 0 \\ 0 & 1 & 0 \\ a-b & 0 & b \\ \end{pmatrix}$$
with $a=rt$ and $b=su$. Also, $rt \in \mathbb{R^*}$ since $r, t \in \mathbb{R^*}$
But! This is where it doesn't make sense to me:
I want $rt-su \in \mathbb{R^*}$
But instead I have:
$rt-su \notin \mathbb{R^*}$ because, we can demonstrate:
Let $r=3, t=4, s=2, u=6$, then $rt-su=0 \notin \mathbb{R^*}$
But, it is assumed closure is satisfied. So where am I going wrong?
The definition of $H$ does not require the $(3,1)$-entry to be nonzero: it only requires the $(1,1)$ and $(3,3)$ entries to be nonzero. The identity matrix lies in $H$ (with $a=b=1$), but it has $(3,1)$-entry equal to $0$; as does any matrix with $a=b\neq 0$.