Denote $\varphi(f(0)) = c$. 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
$$f(\varphi(d) + f(y + c)) = y + c.$$
Note that $f$ is an injective, it implies that
$$\varphi(d) + f(y) + c = \varphi(d) + f(y + c) = \varphi(d) + f(y + c) = f(y),$$
which means
$$\varphi(d) + c = 0.$$
On the other hand, because $\varphi(x) \ge 0$ 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
$$f(\varphi(x)) = \varphi(f(x))$$
but it is clear that $\varphi(x) \ge 0$ for all $x$, which leads to $f(t) \ge 0, \forall t \ge 0$. Replacing $y$ by $f(y)$ and $\varphi(x) = t \ge 0$ 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 $x$. $\square$