Any map $f \colon \mathbb{R} \to \mathbb{R}$ induces a "composition map" $$f^\circ\colon \mathbb{R} \times \mathbb{N} \to \mathbb{R},$$

where $$f^{\circ n}(x) = \underbrace{f \circ \dotsb \circ f}_{n \textrm{ times}} (x).$$

If $f$ happens to have an inverse, the domain of $f^\circ$ can be extended to $\mathbb{R} \times \mathbb{Z}$. Here's my question: is there a "nice way" to further extend $f^\circ$ to have domain $\mathbb{R} \times [0, \infty)$ in general, or $\mathbb{R} \times \mathbb{R}$ if $f$ is invertible?

I want to leave the "nice way" intentionally vague to allow for a variety of answers. But two constraints on this. Obviously one can stitch together something contrived, so "a nice way" should have an element of naturality to it. And in the best case scenario the answer would be a function $f^\circ \colon \mathbb{R}\times \mathbb{R} \to \mathbb{R}$ that, under certain reasonable restrictions (e.g. logarithmic convexity as used in the Bohr-Mollerup Theorem), uniquely satisfies

- $f^{\circ 0} = \textrm{id}$, and
- $f^{\circ (t+1)} = f \circ f^{\circ t}$ for all $t \in \mathbb{R}$.

I'm curious if there's a general answer for arbitrary smooth / analytic $f$, but specifically would like to know the answer for $f(x) = e^x$.