TST

Denote $\phi (f(0))=c$. Clearly, $f$ is a bijective because $$f(f(y))=y+c.$$ Replacing $y$ by $f(y)$ in the relation, we have $$f(y+c)=f(y)+c.$$ Because $f$ is a bijective then there exists a real number $d$ that $f(d)=0$. Replacing $(x,y)$ by $(d,y+c)$, we get $$f(\phi (d)+f(y+c))=y+c.$$ Note that $f$ is an injective, it implies that $$\phi (d)+f(y)+c=\phi (d)+f(y+c)=\phi (d)+f(y+c)=f(y),$$ which means $$\phi (d)+c=0.$$ On the other hand, because $\phi (x)\ge 0$ and the equality only happens when $x=0$, thus $$f(0)=d=0.$$

Hence, $f(f(y))=y$ and replace $y=0$ in the original equation, we get $$f(\phi (x))=\phi (f(x))$$ but it is clear that $\phi (x)\ge 0$ for all $x$, which leads to $f(t)\ge 0,\forall t\ge 0$. Replacing $y$ by $f(y)$ and $\phi (x)=t\ge 0$ for any arbitrary $t\ge 0$, we obtain $$f(y+t)=f(y)+f(t),\forall t\ge 0.$$ Therefore, for all pairs $(x,y)\in \mathbb{R}\times \mathbb{R}$ and $t\ge \max (-y,0)$, we have $$f(x+y)+f(t)=f(x+y+t)=f(x)+f(y+t)=f(x)+f(y)+f(t)$$ or $f$ is additive. We also have proved that $f(x)\ge 0$ for all real numbers $x$, it implies that $f(x)=kx$ for all $x\in \mathbb{R}$. By replacing $f(x)=kx$ in the original equation, we can find that $k=1$. Thus, $f(x)=x$ for all real numbers $x$. $\square$