Suppose $f$ and $g$ are functions that fail to be one-to-one, but $f+g$ is one-to-one. Has anyone ever seen the notation $(f+g)^{-1}$ for the inverse function in that situation? (I find myself wondering if I'm comfortable with this.)
(Example: Consider the six familiar trigonometric function as mappings from $\mathbb R \bmod 2\pi$ to $\mathbb R\cup\{\infty\}$, where the codomain is just the one-point compactification, so that instead of $+\infty$ and $-\infty$ we have one $\infty$ at both ends of the line. All six fail to be one-to-one. But $\sec+\tan$ and $\sec-\tan$ are one-to-one (and onto if we remove the one removable discontinuity from each).)
If $h$ is an invertible function, then $h^{-1}$ can be used to denote its inverse. In particular, we can take $h=f+g$ here. It is not relevant if $f$ or $g$ are invertible. I would not hesitate to use this notation myself, though I can't be sure if I have come across it.
In a similar way, we write $\frac{1}{1+0}=1$, even though $\frac{1}{0}$ is undefined in the real numbers!