ca.analysis and odes - Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$?

Yes, there exist nonlinear solutions.

Multiplying by $e^{x+1}$ and setting $g(x):=e^x f(x)$ transforms the question into finding a solution to $g(x+1)=eg'(x)$ not of the form $e^x(ax+b)$.

Start with any $C^infty$ function on $mathbb{R}$ whose Taylor series centered at $0$ and $1$ are identically $0$, but which is nonzero somewhere inside $(0,1)$. Restrict it to $[0,1]$. Let $g(x)$ on $[0,1]$ be this. Using $g(x+1):=eg'(x)$ for $x in [0,1]$ extends $g(x)$ to a $C^infty$ function $g(x)$ on $[0,2]$, which can then be extended to $[0,3]$, and so on. In the other direction, use $g(x) := int_0^x e^{-1} g(t+1) dt$ to define $g(x)$ for $x in [-1,0]$, and then for $x in [-2,-1]$, and so on. These piece together to give a $C^infty$ function $g(x)$ on all of $mathbb{R}$. The corresponding $f(x)$ satisfies $f(0)=0$ and $f(1)=0$ but is not identically $0$, so it is not linear.