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.
