Slightly generalizing, we may state conditions in terms of $G(x,y):=g(x,h(y))$. For instance:

Let $G$ be a continuous function on $\mathbb{R}^2$. Assume that the equation $G(x,x)=0$ has a unique solution $x^*$ and that for some $M$ any solution of $G(x,y)=0$ verifies $y\le x\le M$. Then any sequence $(x_i)$ such that $G(x_{i+1}, x_i)=0$ does converge to $x^*$.

Indeed by assumption the sequence $x_i$ must be increasing and bounded, thus convergent, and by continuity of $G$, its limit is a solution of $G(x,x)=0$, hence it's $x^*$.

Slightly generalizing, we may state conditions in terms of $G(x,y):=g(x,h(y))$. For instance:

Let $G$ be a continuous function on $\mathbb{R}^2$. Assume that the equation $G(x,x)=0$ has a unique solution $x^*$ and that for some $M$ any solution of $G(x,y)=0$ verifies $y\le x\le M$. Then any sequence $(x_i)$ such that $G(x_{i+1}, x_i)=0$ does converge to $x^*$.

Indeed by assumption the sequence $x_i$ must be increasing and bounded, thus convergent, and by continuity of $G$, its limit is a solution of $G(x,x)=0$, hence it's $x^*$.

Here is a variant producing oscillating sequences.

Let $G$ be a continuous function on $\mathbb{R}^2$. Assume that the system $G(x,y)=G(y,x)=0$ has a unique solution (necessarily with $x=y:=x^*$, otherwise $(y,x)$ would be a second solution). Assume further that any solution to the system $G(x,y)=G(y,z)=0$ verifies either $ z \le x\le y$ or $ y\le x\le z$. Then any sequence $(x_i)$ such that $G(x_{i+1}, x_i)=0$ does converge to $x^*$.

Indeed it follows by induction from the assumptions that any such sequence $(x_i)$ either verifies $$x_0\le x_2\le x_4\le \dots \le x_5\le x_3\le x_1 $$ or $$x_1\le x_3\le x_5\le \dots \le x_4\le x_2\le x_0; $$ in any case, the subsequences $x_{2i}$ and $x_{2i+1}$ converge respectively to $x$ and $y$ solving $G(x,y)=G(y,x)=0$, hence $x=y=x^*$, so the whole sequence $(x_i)$ converges to $x^*$.

Slightly generalizing, we may state reasonable conditions in terms of $G(x,y):=g(x,h(y))$ which implies boundedness of the sequence $(x_i)$, that is all we need. For instance:

If $G$ is continuous and $G(x,x)=0$ has a unique root $x^*$, then any sequence $(x_i)$ such that $G(x_{i+1}, x_i)=0$ does converge to $x^*$, provided $G$ also satisfies: there is $M$ such that $G(x,y)\neq0$ for any pair $(x,y)$ such that either $\min(x,y)> M$ or $\max(x,y)< -M$.

Indeed, in such a situation, one first observe that the sequence $x_i$ must be bounded (there can't be a subsequence $x_{i_j}$ converging to $\pm\infty$). So by compactness any subsequence of $x_i$ possesses a converging sub-sub-sequence. By continuity of $G$, the limit is a solution of $G(x,x)=0$, hence it's $x^*$, and this implies that the sequence $(x_i)$ itself converges to $x^*$.

Slightly generalizing, we may state conditions in terms of $G(x,y):=g(x,h(y))$. For instance:

Let $G$ be a continuous function on $\mathbb{R}^2$. Assume that the equation $G(x,x)=0$ has a unique solution $x^*$ and that for some $M$ any solution of $G(x,y)=0$ verifies $y\le x\le M$. Then any sequence $(x_i)$ such that $G(x_{i+1}, x_i)=0$ does converge to $x^*$.

Indeed by assumption the sequence $x_i$ must be increasing and bounded, thus convergent, and by continuity of $G$, its limit is a solution of $G(x,x)=0$, hence it's $x^*$.