Functions $f$ such that $f(x+1)-f(x-1)=2f'(x)$.

Let $f(x)$ be any differentiable function defined on $[0,1]$ and let $g(x)$ be any integrable function defined on $[1,2]$. Extend $f$ for $x\in (1,2]$ by defining $$f(x) = \int_1^x g(t)\,dt + c$$ where $c$ is chosen to make $f$ continuous at $1$. Also require that $g(1) = f'(1)$, so that $f$ will be also differentiable at $1$. Extend $f$ to $x \in (2,3]$ by $$f(x) = f(x-2) + g(x-1).$$ Now $f$ is defined on $[0,3]$ and satisfies the formula given in the question.

Repeat the above starting with $f$ so far defined on $[0,3]$ to extend the definition of $f$ to $[0,4]$, etc. to extend the definition to $[0, \infty)$.

One can work this procedure backwards to extend the definition of $f$ to all of $\mathbb{R}$. Thus any such $f$ satisfying the original formula can be created from starting functions $f$ and $g$ as described.

All such functions can be obtained in the following way. Start with any smooth ($C^\infty$) function $f:[-1,1]\to\mathbb{R}$ such that $$f^{(n)}(1)-f^{(n)}(-1)=2f^{(n+1)}(0)$$ for all $n\geq 0$. We can then extend $f$ smoothly to $[-1,2]$ by defining $f(x)=f(x-2)+2f'(x-1)$ for $x\in[1,2]$. The equations above guarantee that all the higher derivatives of $f$ at $1$ from the left and right will agree, so $f$ is smooth at $1$. But now, $f$ satisfies $f^{(n)}(x+1)-f^{(n)}(x-1)=2f^{(n+1)}(x)$ for all $n$ and all $x\in [0,1]$. In particular, these equations for $x=1$ allow us to extend $f$ to $[2,3]$ in the same way. We can similarly extend to $[3,4]$ and $[4,5]$ and so on, and also to $[-2,-1]$, $[-3,-2]$, and so on, and obtain a smooth function $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x+1)-f(x-1)=2f'(x)$ for all $x$.

Conversely, given any $f:\mathbb{R}\to\mathbb{R}$ such that $f(x+1)-f(x-1)=2f'(x)$ for all $x$, note that $f$ must be smooth: it is differentiable by assumption, but then its derivative can be written in terms of $f$ itself so its derivative is differentiable, and so on. So restricting $f$ to $[-1,1]$ we get an initial function as above, and the values of $f$ on all of $\mathbb{R}$ can be recovered by the process above.

Let me finally remark that there are indeed many functions $f:[-1,1]\to\mathbb{R}$ satisfying the conditions above. Indeed, those conditions only involve the derivatives of $f$ at $-1$, $0$, and $1$. So we can first prescribe $f$'s behavior near those three points, and then fill in the values between them however we want, as long as we do so smoothly. Moreover, given any two sequences $(a_n)$ and $(b_n)$ of real numbers, if we define $c_n=a_n+2b_{n+1}$, then it is possible to construct $f$ such that $f^{(n)}(-1)=a_n$, $f^{(n)}(0)=b_n$, and $f^{(n)}(1)=c_n$ (by Borel's lemma).