Can $18$ consecutive integers be separated into two groups,such that their product is equal?

This is impossible.

At most one of the integers can be divisible by $19$. If there is such an integer, then one group will contain it and the other one will not. The first product is then divisible by $19$ whereas the second is not (since $19$ is prime) --- a contradiction.

So if this possible, the remainders of the numbers after division by $19$ must be precisely $1,2,3,\cdots,18$.

Now let $x$ be the product of the numbers in one of the groups. Then

$x^2 \equiv 18! \equiv -1 \pmod{19}$

by Wilson's Theorem. However $-1$ is not a quadratic residue mod $19$, because the only possible squares mod $19$ are $1,4,9,16,6,17,11,7,5$.

If $18$ consecutive positive integers could be separated into two groups with equal products, then the product of all $18$ integers would be a perfect square. However, the product of two or more consecutive positive integers can never be a perfect square, according to a famous theorem of P. Erdős, Note on products of consecutive integers, J. London Math. Soc. 14 (1939), 194-198.