Let \( f(x, y) = x + y + xy \). It can be rewritten as \( f(x, y) = (x + 1)(y + 1) - 1 \). Now, consider an arbitrary set of integers \( A = \{a_1, a_2, \dots, a_n\} \). Applying the algorithm mentioned in the problem statement to two integers \( a_i \) and \( a_j \) chosen randomly from the set such that \( i \ne j \), we obtain, \( f(a_i, a_j) = (a_i + 1)(b_i + 1) - 1 \). When this expression is used in subsequent calls to \( f(x, y) \), the \( -1 \) drops, the new number is included in the product and the \( -1 \) gets added again. For instance, using this result and a different number \( a_k \) chosen randomly from the set, we obtain \( f\bigl((a_i + 1)(a_j + 1) - 1, a_k\bigr) = (a_i + 1)(a_j + 1)(a_k + 1) - 1 \). Inductively, this pattern holds, and the algorithm returns a product of all integers in the set minus \( 1 \).

Therefore, the result of the algorithm, when applied to the set \( A \) is \( \prod_{i=1}^n (a_i + 1) - 1 \). If \( A = \{1, 2, \dots, 9\} \), on applying the algorithm, we obtain \( (1 + 1)(2 + 1) \dots (9 + 1) - 1 = 10! - 1 = 3628799 \).