Here is a natural generalization. Let $ A:=\begin{pmatrix}a&b\\ c&d\end{pmatrix} $ be an invertible real matrix, and let $g$ be the associated fractional linear transformation: $ gx=\frac{ax+b}{cx+d}\quad. $ What is natural is to let $g$ acts on the projective line $\mathbb R\cup\{\infty\}$, endowed with the usual topology.
What is the dynamic of the iterations of $g$?
By conjugating $A$, it suffices to consider the cases listed below (corresponding to the possible Jordan forms). In each case, the dynamic is clear. Here is the list:
$\bullet\ \ gx=a\,x, |a| > 1$: the fixed points are $0$, which is repulsive, and $\infty$, which is attractive;
$\bullet\ \ gx=x+1$: the fixed point is $\infty$;
$\bullet\ \ A$ is a rotation by an angle $\theta$. In this case it is better to identify the projective line to a circle in the plane. Then $g$ acts by a rotation of angle $2\theta$.
EDIT A. Here is a consequence of the above observations.
Let $(x_n)_{n\ge0}$ be a sequence of real numbers such that $ x_{n+1}=\frac{ax_n+b}{cx_n+d} $ for all $n$, where $ A:=\begin{pmatrix}a&b\\ c&d\end{pmatrix} $ is an invertible real matrix. Assume also $x_1\neq x_0$.
Then $(x_n)$ converges in $\mathbb R\cup\{\infty\}$ if and only if the eigenvalues of $A$ are real and the trace of $A$ is nonzero.
This condition is satisfied in your case.
EDIT B. Didier Piau asked very kindly for clarification.
We topologize $X:=\mathbb R\cup\{\infty\}$ by adding to the open sets of $\mathbb R$ the complements (in $X$) of the compact sets of $\mathbb R$. Then $X$ is homeomorphic to a circle.
We define the projective line $Y$ as the set of lines through the origin of $\mathbb R^2$.
We consider the following bijection from $Y$ to $X$: we attach $x/y$ to the line through $(x,y)\neq(0,0)$, with the convention $x/y=\infty$ if $y=0$. We identify $X$ and $Y$ via this bijection.
We let G' be the group of invertible two by two real matrices, which acts (by homeomorhisms) on $Y$ in the obvious way. The kernel of this action is the subgroup of nonzero scalar matrices. We denote the quotient group by $G$, and note that the action of $A$ as above on $X$ is given by $ Ax=\frac{ax+b}{cx+d}\quad. $
"Conjugation" means "conjugation in $G$", which is related in a very simple way to conjugation in $G$.
As an example, assume that $A$ is diagonal, i.e. $A=\text{diag}(a,d)$. This gives $ Ax=\frac{a}{d}\ x. $ But $A$ is conjugate (in $G'$) to $\text{diag}(d,a)$. This shows that $x\mapsto\lambda x$ is conjugate (in $G$) to $x\mapsto x/\lambda$. This comes down to switching $0$ and $\infty$ by $x\mapsto 1/x$.
Thank you to Didier for his interest!