It follows from (i) that $f$ is strictly increasing function. Also, (ii) implies
$$x + y \geq f(f(x + y)). \qquad (1)$$

Furthermore, (i) gives $f(f(x+y)) \ge f(f(x)+y)$ and the substitution $x \to y$ and $y \to f(x)$ in (i) implies $f(f(x)+y) \ge f(x)+f(y)$.
Since $f$ is increasing we have $\lim_{x \to 0^+} f(x) = \inf_{x>0} f(x) = \ell \ge 0$. Note that (ii) implies $\lim_{x \to 0^+} f(f(x)) = 0$.
Assume $\ell > 0$. Since $f$ is increasing we have $f(f(x)) \ge f(\ell) > 0$, a contradiction to $\lim_{x \to 0^+} f(f(x)) = 0$. Therefore $\ell = 0$ and $\lim_{x \to 0^+} f(x) = 0$. Letting $y \to 0^+$ in (1) we obtain $x \ge f(x)$ for all positive $x$. It follows now from (i) that
$$x + y \ge f(x + y) \ge f(x) + y, \\ x - f(x) \ge f(x + y) - f(x) - y \ge 0.$$
Fix $x+y$ and let $x \to 0^+$ in the above inequalities. We have $f(x+y) = x+y$ meaning that $f(x) = x$ for all positive $x$. This function is obviously a solution to the problem.