Long comment addressing the triangular function: as fedja remarks in his comment,  when $p(x)$ is supported in $[0,1]$ and symmetric around 1/2, there is a solution from $\infty$ and a solution from $-\infty$ and he thinks this will not join coincide.   The condition that they coincide is  that  $1-g^{-1}(1-x)$ also satisfies  (1) so  $g^{-1}(x) = 1-g(1-x)$ and that $$\epsilon p(x) = g(x) + g(1-x) -1   (2)$$.  With the original functional equation, this allows you to generate a few values of g, it doesn't always get you very far, but what it does generate is perfectly nice looking, imo.  Starting from $x=1/2$ (2)  gives $g(1/2) = (1 + \epsilon p(1/2))/2$.  Using (1) you now know the value of g at x=g(1/2), and the using 2 at x = 1-g(1/2).  Some values of $\epsilon$ e.g., .25, only generate a few values, but here is a picture of a few iterations of this with $\epsilon = .199876$ and the triangular function.


[![picture of same][1]][1]


  [1]: https://i.sstatic.net/Fxcon.png