Solution:
For an arbitrary real number $z$ we set $a=f(z)$, $b=z+f(0)$, $n=f\left(\frac{a+b}{2}\right)$ and $m=f(0)-\frac{a+b}{2}$. Since $\mathbb{D}_f=\mathbb{R}$, $a$, $b$, $n$ and $m$ are well-defined. Substituting $x=0$, $y=z$ into $(*)$ gives $f(z+f(0))=f(f(z))$, hence $f(a)=f(b)$ (I).

Substituting $x=m$ and $y=a$ resp. $y=b$ into $(*)$ gives
$$\begin{array}{l} f(f(m)+a)=2 m+f(f(a)-m) \\ f(f(m)+b)=2 m+f(f(b)-m) \end{array}$$
which, because of (I), we can combine into $f(f(m)+a)=f(f(m)+b)$ (II).

Substituting $x=a$ resp. $x=b$ and $y=n$ into $(*)$ gives
$$\begin{array}{l} f(f(a)+n)=2 a+f(f(n)-a) \\ f(f(b)+n)=2 b+f(f(n)-b) \end{array}$$
which, because of (I), we can combine into $2 a+f(f(n)-a)=2 b+f(f(n)-b)$ (III).

Substituting $x=\frac{a+b}{2}$, $y=0$ into $(*)$ gives $f\left(f\left(\frac{a+b}{2}\right)\right)=a+b+f\left(f(0)-\frac{a+b}{2}\right)$, hence $f(n)=a+b+f(m)$, or, stated differently, $f(n)-a=f(m)+b$ resp. $f(n)-b=f(m)+a$. This together with (III) immediately gives $2 a+f(f(m)+b)=2 b+f(f(m)+a)$, from which, using (II), we now obtain $a=b$, hence $f(z)=z+f(0)$.

Thus it is shown that every function $f$ satisfying $(*)$ must have the form $f(x)=x+c$ ($c=$ constant). Substituting into $(*)$ confirms that every $f$ of this form indeed satisfies the given equation: $f(f(x)+y)=x+y+2 c=2 x+f(f(y)-x)$.