Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
\[f(x) f(y) - f(xy) = x + y\]for all real numbers $x$ and $y.$

Let $n$ be the number of possible values of $f(2),$ and let $s$ be the sum of all possible values of $f(2).$  Find $n \times s.$
Setting $x = y = 1,$ we get
\[f(1)^2 - f(1) = 2,\]so $f(1)^2 - f(1) - 2 = 0.$  This factors as $(f(1) + 1)(f(1) - 2) = 0,$ so $f(1) = -1$ or $f(1) = 2.$

Setting $y = 1,$ we get
\[f(x) f(1) - f(x) = x + 1\]for all $x.$  Then $f(x) (f(1) - 1) = x + 1.$  Since $f(1) \neq 1,$ we can write
\[f(x) = \frac{x + 1}{f(1) - 1}.\]If $f(1) = -1,$ then
\[f(x) = \frac{x + 1}{-2},\]and we can check that this function does not work.

If $f(1) = 2,$ then
\[f(x) = x + 1\]and we can check that this function works.

Therefore, $n = 1$ and $s = 3,$ so $n \times s = \boxed{3}.$