$\begingroup$

A general method for such problems is to use power series with undetermined coefficients. We decide to expand the function $f(x)$ as a power series around $x=0$. The equation to solve for $f$ is $$ f(x) - f^{-1}(x) = e^x-1. \tag1 $$ If $x=0$ then the right side is $0$ and thus $f(0)=f^{-1}(0).$ We are given that $f$ is increasing, thus $f(0)=0.$ Our Ansatz now becomes $$ f(x) = a_1 x + a_2 x^2 + a_3 x^3 + \cdots,\quad g(x) := f^{-1}(x) = b_1 x + b_2 x^2 + b_3 x^3 + \cdots. \tag2$$ We can find the power series for $g(x)$ using the Lagrange inversion theorem or by using the identity $f(g(x)) = x$ and solving for the coefficients of $g(x).$ By substituting equation $(2)$ into equation $(1)$ using the expansions of $f$ and $g$ we can solve for $a_1,a_2,a_3,\dots.$ The equation to solve for $a_1$ is $a_1-1/a_1=1$ whose positive solution is $a_1 = \phi := (1+\sqrt{5})/2.$ The rest of the coefficients are solutions of linear equations and the result is $$ f(x) = \phi\, x + \frac{\phi}4 x^2 + \frac{1+7\phi}{72} x^3 + \frac{13+170\phi}{6336} x^4 + \frac{-1279+5003\phi}{950400} x^5 + \cdots. \tag3 $$ There seems to be no obvious pattern to the coefficients. Of course, the radius of convergence is not yet known, and also if the power series is increasing. This is one of the limitations of this method. If the function has a power series, then we can find it, but we don't know much else about the function.

A general method which proves existence uses ideas from a solution to the Grossman's Constant problem. Specifically, the solution by Gabor Nyerges from 2000 in his "The Solution of the Functional Equation $x = (1+F(x))F^2(x)$" available from the Internet Archive Wayback Machine. His solution can be easily generalized to the functional equation $(1)$.

In more detail, consider the Fibonacci sequence $F_n$. The forward recurrence is $\,F_{n+2} = F_{n+1} + F_n\,$ for all integer $n.$ The backward recurrence is $\,F_{n-2} = F_n - F_{n-1}.\,$ There are two linearly independent solutions to the recurrence $\,u_{n+2} = u_{n+1} + u_n.\,$ One for each root of the characteristic polynomial $\,x^2 - x - 1.\,$ One root is the golden ratio $\,\phi := (1+\sqrt{5})/2 > 1\,$ the other is $\,0 > -1/\phi > -1.\,$ The solution for $\phi$ is exponentially increasing as $\,n \to \infty\,$ and also exponentially decreasing to $0$ as $n \to -\infty.\,$ The solution for the other root is exponentially alternating decreasing to $0$ as $\,n \to \infty\,$ and also exponentially alternating increasing as $\,n \to -\infty.\,$ Given two initial values $a_0$ and $a_1,$ the sequence is uniquely determined for all integer $n$ and is a linear combination of the two exponential solutions. As $\,n \to -\infty\,$ and if $\,u_0\,\phi = u_1,\,$ then $u_n$ decreases exponentially to $0,$ but if $\,u_0\,\phi

e u_1,\,$ then the other solution is dominant and $u_n$ will eventually alternate in sign and become unbounded.

A similar situation arises in this question. Let us generalize the problem. Suppose that we want to construct $f(x)$ which is an increasing bijection on the reals that satisfies $$ f(x) - f^{-1}(x) = s(x) \tag{4} $$ where $\,s(x)\,$ is a differentiable function that satisfies $$ s(0) = 0 \quad \text{ and } \quad s'(x) > 0 \;\forall x. \tag{5} $$ Given $\,0 < x < y\,$ construct a sequence $\,u_n\,$ such that $$ u_0 = x,\quad u_1 = y,\quad u_{n+2} = s(u_{n+1}) + u_n\;\; \forall n\in\mathbb{Z}.\tag{6} $$ Notice that this is the Fibonacci recurrence if $\,s(x) = x.\,$ Notice that $\,u_{-1} = u_1 - s(u_0),\;\;$ $u_{-2} = u_0 - s(u_{-1}),\,$ and so on where we stop the backwards recursion if $\,u_n\,$ ever goes negative. Notice also that $\,u_{2k}\,$ and $\,u_{2k+1}\,$ are increasing sequences iff $\,u_n > 0\;\; \forall n \in\mathbb{Z},\,$ but the sequence $\,u_n\,$ itself may not be. We need to find out exactly when it is monotone increasing.

Now suppose that we know $\,s(x) = c_1\,x + O(x^2)\,$ where $\,c_1 > 1.\,$ By using arguments similar to the Fibonacci recurrence, we find that given $x$ there is a unique value of $\,y\,$ such that the sequence $\,u_n\,$ is such that $\,u_n \to 0\,$ monotonically as $\,n \to -\infty.\,$ Now define $\,f(x)\,$ to be this unique $\,y.\,$ This implies that $\, u_{n+1} = f(u_n)\;\; \forall n\in\mathbb{Z}.\,$ A similar argument holds for $\,x < 0\,$ using the forward recurrence. Thus $\,f(x)\,$ is defined for all reals. Because of equations $(5)$ and $(6)$ it is monotone increasing. Suppose that $\,s(x)\, > L > 0\,$ for $\,x > K > 0\,$ for some $\,L,K\,$ and the same with $>$ replaced with $<$. Then $\,f(x)\,$ is unbounded and hence it is a bijection. Standard $\delta-\epsilon$ methods can prove that $\,f(x)\,$ is differentiable.

Notice that similar arguments construct $\,f(x)\,$ if it is required to be a decreasing function instead of increasing. Notice that if $\,s(x) = c_1\,x\;\; \forall x\in\mathbb{R}\,$ then we have the linear recursion case and $\,f(x) = a_1\,x\,$ where $\,a_1\,$ is a positive root of the characteristic polynomial $\, x^2 - c_1\,x - 1.$