For $x=y-1-f(y)$ we get $f(2y-2f(y)-1)=2f(y)-2y-1$ for all real $y$. Now take $y=2z-2f(z)-1$ and get $f(4z-4f(z)-1)=4f(z)-4z-1$ for all real $z$. Inductively we get $f(2^n(x-f(x))-1)=2^n(f(x)-x)-1$ for all real $x$ and natural number $n$. Now we shall prove that $g(x)=f(x)-x=0$ for all real numbers $x$. Rewrite the last equation in terms of $g(x)$ and obtain $g(-2^ng(x)-1)=2^{n+1}g(x)$ for all real $x.$ Hence we get $\frac{g(-2^ng(x)-1)}{-2^ng(x)-1} = \frac{2^{n+1}g(x)}{-2^ng(x)-1}$ Since $\frac{g(x)}{x}$ has finitely many values for non-zero reals $x$ the function $h(n, x)=\frac{2^{n+1}g(x)}{2^ng(x)+1}$ has finitely many values. Now assume that $g(x_0) \neq 0$ for some real number $x_0$. Then $2-h(n, x_0)=\frac{2}{2^ng(x_0)+1}$ cannot take finitely many values since for distinct $n$ values, say $n_1, n_2, \: 2^{n_1}g(x_0) \neq 2^{n_2}g(x_0)$ since $g(x_0) \neq 0.$ So, the only such function is $f(x)=x$ for all real numbers $x.$