When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?

This is an example of a delay differential equation, and it has a very large number of solutions. I will begin by discussing solutions $f\colon[0,\infty)\to\mathbb{R}$, and then extend to solutions on the real line.

Suppose first that we have defined $f$ on the interval $[0,1]$, say $f(x) = f_0(x)$ for some differentiable function $f_0\colon[0,1]\to\mathbb{R}$. For extending solutions forwards, the best way to write the equation is $$ f(x+1) \;=\; f(x) + f'(x). $$ Using this equation, it is not hard to define $f$ on the interval $[1,2]$. Specifically, if $x\in[1,2]$, then $$ f(x) \;=\; f_0(x-1) + f_0'(x-1). $$ Note that this function will only be differentiable if $f_0$ was twice differentiable on $[0,1]$. Moreover, the derivatives at $1$ will only match up if $f_0'(1) = f_0'(0) + f_0''(0)$. Assuming this is the case, it is easy to extend again to $[2,3]$. Specifically, if $x\in[2,3]$, then $$ f(x) \;=\; f(x-1) + f'(x-1) \;=\; f_0(x-2) + 2 f_0'(x-2) + f_0''(x-2). $$ In general, assuming that:

1. The function $f_0$ is $C^\infty$ on $[0,1]$ and,
2. $f_0^{(k)}(1) = f_0^{(k)}(0) + f_0^{(k+1)}(0)$ for all $k\geq 0$,

we can extend $f$ to all of $[0,\infty)$ by the formula $$ f(x) \;=\; \sum_{k=0}^{\lfloor x\rfloor} \,\binom{\lfloor x\rfloor}{k}f_0^{(k)}(x-\lfloor x\rfloor) $$

Now let's discuss how to extend backwards. Assuming we have defined $f$ on $[0,\infty)$, the formula for $f$ on $[-1,0]$ is determined by the initial-value problem $$ f'(x) + f(x) \;=\; f_0(x+1),\qquad f(0) = f_0(0). $$ This is a first-order differential equation, so it must have a solution. Moreover, assuming $f_0$ is $C^\infty$ on $[0,1]$, it follows from the Picard-Lindelöf Theorem that there is a unique $C^\infty$ solution for $f$ on $[-1,0]$. Note that the derivatives at $0$ will automatically match, since $f_0(1) - f(0) = f_0'(0)$. Continuing in this fashion, we can extend any solution backwards all the way to $-\infty$.

We conclude that any function $f_0\colon[0,1]\to\mathbb{R}$ satisfying the two conditions listed above extends to a solution on the entire real line.

Edit: Incidentally, there is a simple integral formula for $f$ on $(-\infty,0]$, which can be used to compute values for $f$ on a computer: $$ f(x) \;=\; f(0) + e^{-x-1}\int_{x+1}^1 e^{t}\,f(t)\,dt $$ This can also be used to provide an alternative proof that the function extends backwards, without appealing to existence and uniqueness of ODE's. I found this formula by solving the initial value problem $f(x) + f'(x) = f(x+1)$ using integrating factors.

Let $\lambda = \alpha + \beta i \in \mathbb{C}$ be any root of the equation $\lambda = e^{\lambda} - 1$ subject to $\beta \ge 0$ and $\delta$ any real number, then $$f(x) = \Re( e^{\lambda x + i\delta} ) = e^{\alpha x} \cos(\beta x + \delta)\tag{*}$$ will be a $C^{\infty}$ solution for the functional equation $f'(x) = f(x+1) - f(x)$.

For example, $\lambda = e^{\lambda} -1 $ has a root at $\sim 2.088843015613044 + 7.461489285654254 i$.
This gives us following $C^{\infty}$ solution:

$$f(x) \sim e^{ 2.088843015613044\,x}\,\cos(7.461489285654254\,x)$$

Other solutions can be constructed by forming linear combination of solutions of the form in $(*)$. Please note that $\lambda = e^{\lambda} - 1$ has a double root at $\lambda = 0$, there are two solutions at $\lambda = 0$ which one can use:

$$ \Re( e^{\lambda x} )\Big|_{\lambda = 0} = 1 \quad\quad\text{ and }\quad\quad \frac{\partial}{\partial \lambda} \Re( e^{\lambda x} )\Big|_{\lambda = 0} = x $$

Commenting to achille hui's answer, the idea of choosing $\lambda$ comes from the observation that the translation operator $\tau_{h} f (x) = f(x+h)$ corresponds to the exponential of the differential operator

$$\tau_{h}f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!}h^{n} = e^{h d/dx} f(x).$$

on the space of analytic functions. So writing $D = d/dx$, the given equation can be transformed into

$$ e^{D}f(x) - f(x) = Df(x) \quad \Longleftrightarrow \quad (e^{D} - 1 - D)f(x) = 0.$$

By making an ansatz that a solution $f(x)$ has the form $f(x) = e^{\lambda x}$, it is easy to see that the equation above is equivalent to $e^{\lambda} - 1 - \lambda = 0$, the very starting point of the achille hui's answer.