First, $f$ is bijective because, plugging $x = 1$, we get $f(f(y) + f(1)) = 2f(1) + x$; $x + 2f(1)$ spans all real numbers (so $f$ is surjective) and $f(x) = f(y) \iff f(x) + f(1) = f(y) + f(1) \iff f(f(x) + f(1)) = f(f(y) + f(1)) \iff 2f(1) + x = 2f(1) + y \iff x = y$ (so $f$ is injective).
Choose $x \neq 0$ and $y$ such that $2f(x) + xy = f(x) \iff y = -\frac{f(x)}{x}$. By injectivity,
$$x f\left(-\frac{f(x)}{x}\right) + f(x) = x \iff f\left(-\frac{f(x)}{x}\right) + \frac{f(x)}{x} = 1 \quad (*)$$
Now, set $x = y = 0$: $f(f(0)) = 2f(0)$. Let $a = f(0)$, so that $f(a) = 2a$. If $a = 0$, setting $y = 0$ we get $f(f(x)) = 2f(x)$ and by surjectivity $f(x) = 2x$ for all $x$. This function doesn't work (by testing), so $a$ can't be 0. So $a \neq 0$ and plugging $x = a$ in $(*)$, we obtain (recall that $f(a) = 2a$):
$$f(-2) + 2 = 1 \iff f(-2) = -1$$
Now set $y = 0$: $f(f(x) - x) = 2(f(x) - x)$. Now we find all real numbers $k$ such that $f(k) = 2k$. If there is only one possible value for $k$, we are almost done because $f(x) - x = k \iff f(x) = x + k$ and by testing the function, $f(x) = x + 1$.
Let $m$ be such that $f(m) = k$ (it exists because $f$ is surjective). Plug $x = m$ and $y = 0$: recalling that $f(0) = a$,
$$f(mf(0) + f(m)) = 2f(m) \iff f(ma + k) = 2k = f(k)$$
By injectivity,
$$ma + k = k \iff ma = 0 \iff m = 0$$
because we already proved that $a \neq 0$. So there is only one value for $k$, namely, $k = f(0) = a$. Then, we're done: the only function is $f(x) = x + 1$.