Yes, there exist nonlinear solutions.
Multiplying by ex+1 and setting g(x):=exf(x) transforms the question into finding a solution to g(x+1)=eg′(x) not of the form ex(ax+b).
Start with any Cinfty function on mathbbR 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 xin[0,1] extends g(x) to a Cinfty function g(x) on [0,2], which can then be extended to [0,3], and so on. In the other direction, use g(x):=intx0e−1g(t+1)dt to define g(x) for xin[−1,0], and then for xin[−2,−1], and so on. These piece together to give a Cinfty function g(x) on all of mathbbR. The corresponding f(x) satisfies f(0)=0 and f(1)=0 but is not identically 0, so it is not linear.
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