Solution:

The answer is (C). If $a, b, x, y$ are all greater than or equal to $2$, then $a b \geq a+b = x y \geq x+y = a b$, so every inequality must be an equality and the only possible quadruple is $(2,2,2,2)$.

If instead one of the four numbers is equal to $1$, say for example $b=1$, then substituting the equality $a = x + y$ into $a + 1 = x y$ one obtains $(x-1)(y-1) = 2$, from which $x=2, y=3$ or vice versa. One solution is thus $(a, b, x, y) = (1,5,2,3)$, and the others are obtained from this one by swapping the roles of $a, b$, the roles of $x, y$, or the roles of the pair $(a, b)$ and the pair $(x, y)$: one thus obtains eight other solutions, namely $(1,5,2,3), (1,5,3,2), (5,1,2,3), (5,1,3,2), (2,3,1,5), (2,3,5,1), (3,2,1,5), (3,2,5,1)$.