Substituting $ x_1 = 0$ into (iii) yields $ h(k(0), x_2) = 0$ for all $ x_2 \in S$. But (ii) tells us that $ k(0) = 0$, so $ h(0, x_2) = 0$ for all $ x_2 \in S$. But (i) tells us that $ k(x_2) = h(0, x_2)$ for all $ x_2 \in S$, implying that $ k(x) = 0$ for all $ x \in S$. But then, $ h(0, x_2) = 0 = h(k(x_1), x_2) = x_1$ for all $ x_1 \in S$, implying that $ x_1 = 0$ for all $ x_1 \in S$. In other words, the only possible set $ S$ can be is $ \{0\}$. In addition, the set $ \{0\}$ trivially satisfies the conditions of our problem (by setting $ k(0) = 0$ and $ h(0,0) = 0$), so we are done.