, 2011

$x = y = 0$ yields $f(0) = 0$ and $y = 0$ yields $f(-x) = -f(x)$, $x \in \mathbb{R}$.

$y = 1 \Rightarrow f(2x + 1) = 2f(x) + f(1) \quad (3)$

$y = -1 \Rightarrow f(2x - 1) = 2f(x) - f(1) \quad (4)$

Denote $x * y = xy + x + y$ and notice that (ASOC): $x * (y * z) = (x * y) * z$.
For $x = 1$, $y * 1 = 2y + 1$, hence

$$f(y * 1) = 2f(y) + f(1), \forall y \in \mathbb{R} \quad (5)$$

Therefore
$$\begin{align*} f(x * (y * 1)) &= f(x * (2y + 1)) \\ &\stackrel{(1)}{=} f(x(2y + 1) - x - 2y - 1) + 2f(x) + 2f(2y + 1) \\ &= f(2(xy - y) - 1) + 2f(2y + 1) + 2f(x) \\ &\stackrel{(3,4)}{=} 2f(xy - y) + f(x) + 2f(y) + f(1) \tag{6} \end{align*}$$

and
$$\begin{align} f((x * y) * 1) &\stackrel{(5)}{=} 2f(x * y) + f(1) \stackrel{(1)}{=} 2(2f(x) + 2f(y)) + 2f(xy - x - y) + f(1) \nonumber \\ &= 2f(xy - x - y) + 2f(x) + 2f(y) + f(1). \tag{7} \end{align}$$

Relations (6), (7) and (ASOC) yield:
$$f(xy - x - y) + f(x) = f(xy - y), \forall x, y \in \mathbb{R}.$$
The last relation means that $f(u + v) = f(u) + f(v)$ for all $u, v \in \mathbb{R}$ which can be written in the form $u = x, v = xy - x - y$, that is there exists $x, y \in \mathbb{R}$ so that $x = u$ and $y = \frac{v+u}{u-1}$; this happens if $u \neq 1$ or $u = 1 = -v$.

It remains to check that $f(1 + v) = f(1) + f(v)$, $\forall v \in \mathbb{R}$. Indeed, (4) and $x \mapsto x + 1$ imply
$$f(2x + 1) = 2f(x + 1) - f(1) \stackrel{(3)}{=} 2f(x) + f(1),$$
$$\text{whence } f(x + 1) = f(x) + f(1), \forall x \in \mathbb{R}.$$