UkraineMO

Answer: it doesn't exist.

Let us assume such function exists. Let us substitute $y = 0$ in the given inequality:
$$f(x - f(0)) \le x,$$
Then, substitute $x = x + f(0)$ there:
$$f(x) \le x + f(0). \quad (1)$$
Let us substitute $x = f(y)$ in the initial inequality. We will get:
$$\begin{cases} f(0) \le f(y) - y f(f(y)) \le y + f(0) - y f(f(y)) \text{ or} \\ y f(f(y)) \le y. \end{cases}$$
From the last inequality while $y < 0$ we will have:
$$1 \le f(f(y)) \le f(y) + f(0) \le y + 2f(0). \quad (1)$$
The last inequality has to be satisfied for any $y < 0$, that is surely impossible.