According to conditions (I)-(III) there is a positive integer $x$ and two digits $y \neq z$ s.t. $(1) \;\; a = 10x + y$, $(2) \;\; b = 10x + z$, $(3) \;\; y + z = 10$. Hence we obtain $ab = (10x + y)(10x + z) = 100x^2 + 10x(y + z) + yz = 100x^2 + 10x \cdot 10 + yz$, i.e. $(4) \;\; ab = 100x(x + 1) + yz$. The fact that $0 \leq yz < 100$ (since $y$ and $z$ are digits) combined with formula (4) means the product $yz$ is the two last digits of $ab$. Consequently the hundreds digit $d_2$ of $ab$ is given by the congruence $(5) \;\; d_2 \equiv x(x + 1) \pmod{10}$. Since both 10 and $x(x+1)$ are even integers, the digit $d_2$ is even by congruence (5). q.e.d.