Suppose that $f(0) = a$. Let $x = y = 0$, we have
$$f\left(\frac{a}{2}\right) = f(a) - a$$
Let $x = \frac{a}{2}, y = 0$, we have
$$f\left(\frac{f(a) - \frac{a}{2}}{2}\right) = 2a$$
Let $x = a, y = 0$, we have
$$f\left(\frac{f(a) + a}{2}\right) = 2a$$

Since $f$ is increasing, so for every $\frac{f(a)-a}{2} \le t \le \frac{f(a)+a}{2}$ we have $f(t) = 2a$.
Hence $f(\frac{f(a)+a}{2}) = 2a$. From the other hand by letting $x = y = \frac{a}{2}$, we have
$$f(\frac{f(a)+\frac{a}{2}}{2}) = 2a - f(a) + f(f(a) - a)$$
Hence we have
$$2a = 2a - f(a) + f(f(a) - a) \Rightarrow f(a) = f(f(a) - a)$$
Letting $x = 0, y = \frac{a}{2}$ in the main equation results in
$$f(a) = -a + f(f(a) - a) \Rightarrow -a = 0 \Rightarrow a = 0 \Rightarrow f(0) = 0$$
Now letting $x = 0$ in the main equation results in
$$f(y) = f(f(y))$$
Then let $x = y$, and we have
$$f(\frac{3}{2}x + \frac{1}{2}f(x)) = 2x - f(x) + f(f(x)) = 2x$$
So $f$ is surjective. Hence in the equation $f(y) = f(f(y))$, $f(y)$ can be all values in domain, so $f$ is the identity function.