Solve $f(x+f(2y))=f(x)+f(y)+y$

EDIT: I now have a complete solution.

$$f:\mathbb{R}^+\to\mathbb{R}^+$$ $$f(x+f(2y))=f(x)+f(y)+y$$

Note that $f(a+f(2x_1)+f(2x_2)+...+f(2x_n))=f(a)+f(x_1)+x_1+f(x_2)+x_2+...+f(x_n)+x_n$

Let us calculate the expression $f(x+f(2(y+f(2z)))$ in two ways: $$f(x+f(2(y+f(2z)))=f(x)+f(y+f(2z))+y+f(2z)=f(x)+f(y)+f(z)+z+y+f(2z)$$ $$f(x+f(2(y+f(2z)))=f(x+f(2y+f(2z)+f(2z)))=f(x+f(2y)+2f(z)+2z)=f(x+2z)+f(y)+y+2f(\frac{z}{2})+2\cdot{}\frac{z}{2}$$ Thus, cancelling $f(y)+y+z$ we get $$f(x)+f(z)+f(2z)=f(x+2z)+2f(\frac{z}{2})$$ i.e. (for $y=2z$) $$f(x+y)=f(x)+f(y)+f(\frac{y}{2})-2f(\frac{y}{4})$$

Since the expression $f(x+y)-f(x)-f(y)$ is symmetric in $x$ and $y$, we get that $f(\frac{y}{2})-2f(\frac{y}{4})=f(\frac{x}{2})-2f(\frac{x}{4})$, i.e. $f(2y)-2f(y)=f(2x)-2f(x)$ for all $x,y$. so $f(2x)-2f(x)$ is some constant $-c$. So we obtained: $$f(x+y)=f(x)+f(y)-c$$ Define $i=inf_{\mathbb{R}^+}(f(x))$. Clearly, $0\le{}i\le{}\infty$. as $x$ and $y$ go through all positive real values, the infimum of LHS is $i$ and the infimum of RHS is $2i-c$, therefore $c=i$. since $f(y)-i\ge{}0$, we get $f(x+y)\ge{}f(x)$, so $f$ is monotonically increasing.

Define $g:\mathbb{R}^+\to{}\mathbb{R}^+$ as $g(x)=f(x)-i$. from $f(x+y)=f(x)+f(y)-c$ we get that $g$ is additive, i.e. $g(x+y)=g(x)+g(y)$. Thus, by induction, $g(k)=kg(1)$ for all $k\in{}\mathbb{N}$. This argument can be expanded for all $k\in\mathbb{Q}^+$ (why?), so $g(q)=qg(1)$ for all $q\in{}\mathbb{Q}^+$. but since $f$ is monotonic we get that $g$ is monotonic, so $g(x)=xg(1)$ for all $x>0$. Thus $g$ is linear, so $f$ is linear. Substituting $f(x)=ax+b$ in the original equation yields the solutions: $$a=1$$ $$a=-\frac{1}{2}, b=0$$ We are interested in solutions that give only positive values in the image of $f$. Thus $f(x)=x+b$ for some constant $b\ge{}0$. This function satisfies the equation for all $b\ge{}0$.