Since you are just looking for a pair $f$ and $F$, you don't have to solve a functional equation for $\varphi$ and $\psi$. Take for instance $\varphi(x)=\sin(x)$, $\psi(x)=\cos(x)$. Then $I+\varphi$ and $I+\psi$ are both homeomorphisms $\mathbb R\to\mathbb R$, so you can define $f := \varphi\circ(I +\varphi)^{-1}$ and $F:=\psi\circ(I+\psi)^{-1}$. 

*Rmk* For a smooth example (actually, analytic) you may take $\varphi(x)=a\sin(x)$, $\psi(x)=a\cos(x)$ with $|a|<1$, so that now $I+\varphi$ and $I+\psi$ are smooth diffeos. Of course, if you don't insist that the functions be defined on the whole line, there is even more freedom in the choice of $\varphi$.