Taiwan IMO Selection Camp

The unique solution is $f(x) = x - 1, x \in \mathbb{R}$.
We let $g(x) = f(x) + 1$ and prove that $g(x) = x$ for all real numbers $x$. Then the condition of the problem can be written as
$$g(1+xy)-g(x+y) = (g(x)-1)(g(y)-1) \quad \text{for all } x,y \in \mathbb{R} \text{ with } g(-1) \neq 1. \quad (1)$$
Let $C = g(-1) - 1 \neq 0$. Setting $y = -1$ in (1), we obtain
$$g(1-x) - g(x-1) = C(g(x) - 1). \quad (2)$$
Setting $x = 1$ in (2) gives $C(g(1) - 1) = 0$. Therefore, since $C \neq 0$, we have $g(1) = 1$. Substituting $x = 0$ and $x = 2$ respectively produces $g(0) = 0$ and $g(2) = 2$.
We observe that
$$g(x) + g(2 - x) = 2 \quad \text{for all } x \in \mathbb{R}, \quad (3)$$
$$g(x + 2) - g(x) = 2 \quad \text{for all } x \in \mathbb{R}. \quad (4)$$
Replacing $x$ in (2) by $1 - x$ gives $g(x) - g(-x) = C(g(1-x) - 1)$, and replacing $x$ by $-x$ gives $g(-x) - g(x) = C(g(1+x) - 1)$. Adding the two equations gives $C(g(1-x) + g(1+x) - 2) = 0$. Hence, since $C \neq 0$, we obtain (3).
Let $u, v$ be such that $u + v = 1$. Substituting $(x, y)$ respectively by $(u, v)$ and $(2 - u, 2 - v)$ in (1) gives $g(1 + uv) - g(1) = (g(u) - 1)(g(v) - 1)$, $g(3 + uv) - g(3) =$
$$g(uv + 3) - g(uv + 1) = g(3) - g(1).$$
Every $x \le \frac{5}{4}$ can be expressed in the form $x = uv + 1$ where $u + v = 1$ (since the quadratic polynomial $t^2 - t + (x-1)$ has real roots when $x \le \frac{5}{4}$). Therefore, when $x \le \frac{5}{4}$, $g(x+2)-g(x) = g(3)-g(1)$. Since $g(x) = x$ holds at $x = 0, 1, 2$, letting $x = 0$ gives $g(3) = 3$. This proves that (4) holds when $x \le \frac{5}{4}$. If $x > \frac{5}{4}$ then $-x < \frac{5}{4}$, so $g(2-x)-g(-x) = 2$. On the other hand, from (3) we get $g(x) = 2-g(2-x)$, $g(x+2) = 2-g(-x)$, so that $g(x+2) - g(x) = g(2-x) - g(-x) = 2$. Hence (4) holds for all real numbers $x$.
Replacing $x$ in (3) by $-x$ gives $g(-x) + g(2+x) = 2$. From (4) we obtain $g(x) + g(-x) = 0$ for all $x$. Substituting $(x,y)$ respectively by $(-x,y)$ and $(x,-y)$ in (1) gives $g(1-xy)-g(-x+y) = (g(x)+1)(1-g(y))$, $g(1-xy)-g(x-y) = (1-g(x))(g(y)+1)$. Adding the two equations gives $g(1-xy) = 1-g(x)g(y)$. From (3) we obtain $g(1+xy) = 1+g(x)g(y)$. Therefore, from the original equation (1), we obtain the form $g(x+y) = g(x)+g(y)$, so $g$ is additive.
From additivity we get $g(1+xy) = g(1)+g(xy) = 1+g(xy)$, and since $g(1+xy) = 1+g(x)g(y)$ has already been obtained above, we get $g(xy) = g(x)g(y)$. In particular, setting $y = x$ gives $g(x^2) = g(x)^2 \ge 0$ for all real numbers $x$, meaning that $g(x) \ge 0$ when $x \ge 0$. Since $g$ is additive and bounded below on $[0, +\infty]$, $g$ is linear, and moreover $g(x) = g(1)x = x$ for all real numbers $x$.
In summary, the unique solution is $f(x) = x-1, x \in \mathbb{R}$.