Other answers have established that there is at least one other solution besides the two intended by the original questioner. Let's determine rigorously what all the solutions are.
If any of the numbers is a 2, then we must also have $2\times2=4$ and similarly 8 and 6. Similarly if any of them is an 8. Therefore, if we have a 2 or an 8 then we have {2,4,6,8}.
If any of the numbers is a 3, then we must also have $3\times3=9$, hence 7, hence 1. Similarly if any of them is a 7. Therefore, if we have a 3 or a 7 then we have {1,3,7,9}.
If neither of those conditions holds then the numbers we have are a subset of {1,4,5,6,9}. It is easy to verify that {1,4,6,9} is a solution. Are there others?
If so, they must contain 5 together with exactly three of {4,6,9}. We can't have both 5 and 4, or both 5 and 6, because then we'd need 0 and the numbers have to be positive. So this isn't possible.
Hence: the possible solutions are in fact exactly the ones already found: {1,3,7,9}, {2,4,6,8}, and {1,4,6,9}. (Odd numbers, even numbers, squares.)