If $f : (0,\infty) \rightarrow \mathbb{R}$ and $g:\mathbb{R} \rightarrow (0,\infty)$ such that $f \circ g = I_\mathbb{R}$ where $I_\mathbb{R}$ is the identity function.
Wouldn't a perfect example of this be if $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x}$ thus $f \circ g = x$ as required?
$f \circ g = I_\mathbb{R}$ for any two functions where $f^{-1}(x)=g(x)$, or vice versa, by definition of the inverse function. Also, as $f(x)$ and $g(x)$ are inverse functions, $f \circ g =g\circ f$. Of course, the domain of $f \circ g$ is defined as the intersection of the sets of the codomain of $g(x)$ and the domain of $f(x)$. The domain of $g \circ f$ is the intersection of the codomain of $f(x)$ and the domain of $g(x)$. The domain of $y=x$ is $ \mathbb{R}$, so we need the domain of $f \circ g $ or $g \circ f$ to be the same. Notice too that the codomain of $g(x)$ is the same as the domain of $f(x)$, and the domain of $g(x)$ is the same as the codomain of $f(x)$. Thus, as long as at least one of the 4 domains and codomains is $\mathbb{R}$, that means that either $f \circ g$ or $ g \circ f$ has a domain of $\mathbb{R}$, and thus is equal to $I_\mathbb{R}$. Unfortunately, because neither the domain or codomain of $\frac{1}{x}$ is equal to $\mathbb{R}$, your example does not quite work.