By taking $x = 1$ into (1), we get
$$f(f(y) - f(1)) = y + 2f(1), \quad \forall y \in \mathbb{R}. \quad (2)$$
Hence $f$ is bijection and so, there exists a unique real number $a$ such that $f(a) = 0$. Plugging $x = a$ into (1), we have
$$f(af(y)) = ay, \quad \forall y \in \mathbb{R}. \quad (3)$$
Plugging $y = 0$ into (3), we have $f(af(0)) = 0 = f(a)$. Combined with $f$ is injective, we get $af(0) = a$. Hence, $a = 0$ or $f(0) = 1$.
Consider the case $a = 0$, i.e $f(0) = 0$. Plugging $y = 0$ into (1), we have $f(-f(x)) = 2f(x)$. Since $f$ is surjective, we conclude that $f(x) = -2x$ for all $x \in \mathbb{R}$. But this function is not satisfying the equation (1). Hence, $a \neq 0$ and we get $f(0) = 1$.
Plugging $x = 0$ into (1), we have $f(-1) = 2$. Plugging $y = a$ into (3), we have $a^2 = f(0) = 1$, i.e $a = 1$ (because $f(-1) = 2$), i.e $f(1) = 0$.
Since $f(1) = 1$, we can write the equation (2) as the form
$$f(f(y)) = y, \quad \forall y \in \mathbb{R}. \quad (2')$$
By plugging $f(y)$ instead of $y$ into (1) and using (2'), we get
$$f(xy - f(x)) = 2f(x) + xf(y), \quad \forall x, y \in \mathbb{R}.$$
In this equation, consider $x \neq 0$ and putting $y = \frac{f(x)}{x}$, we have
$$1 = 2f(x) + x f\left(\frac{f(x)}{x}\right),$$
hence
$$f\left(\frac{f(x)}{x}\right) = \frac{1 - 2f(x)}{x}, \quad \forall x \neq 0.$$
By putting $y = \frac{f(x)}{x}$ into (1) and using the above result, we have
$$f(1 - 3f(x)) = 3f(x), \quad \forall x \neq 0.$$
Note that $f$ is bijective and $f(0) = 1$, so for all $x \neq 0$, we can see that $1 - 3f(x)$ can get all real values, except $-2$. Therefore, from the above result, we conclude that $f(x) = -x + 1$ for all $x \neq -2$.
In particular, $f(3) = -2$. Plugging $y = 3$ into (2'), we have $f(-2) = 3$. Therefore, $f(x) = -x + 1$ for all $x \in \mathbb{R}$. It is easy to see that this function satisfying the condition. ■