You’re looking for $P(n,k)$, the number of partitions of $n$ into $m$ parts; there is no nice closed expression for this number. Formula $(59)$ here gives the recurrence
$$P(n,m)=P(n-1,m-1)+P(n-m,m)\;.$$
This isn’t hard to derive. If a partition of $n$ into $m$ parts has a part of size $1$, it comes from a partition of $n-1$ into $m-1$ parts by adding a part of size $1$. Otherwise, it comes from a partition of $n-m$ into $m$ parts by adding $1$ to each part. The initial conditions are $P(n,m)=0$ for $m>n$, $P(n,n)=1$, and $P(n,0)=0$ for $n>0$.
See the two links and their references for more information.
Leaving aside the product restriction for now, you can consider the ways to split each number separately. With four twos, you have five choices for how many are on the left. With two threes, you have tree choices for how many are on the left. This gives $5 \cdot 3=15$ choices, but we subtract two because all the numbers on the same side is not allowed. We have counted each division twice except for those which have exactly half the numbers on each side. That is possible here. It is exactly the case where the products will be equal. We delete that case and divide by two to get the six cases you found. If you have $a_1$ of the first number, $a_2$ of the second, etc. there will be $$\frac 12\left(\prod (a_i+1)[-1]\right)-1$$ where the $[-1]$ applies if all the $a_i$ are even and the last $-1$ removes the case where one group is even.
This will be fine as long as there is no way to make a case with matching products unless we split the numbers evenly, such as the numbers being distinct primes. If your list were $[2,2,2,2,4,4]$ we would have other combinations to exclude and you would just have to compute.
Best Answer
One way to do such a split up is to fill an $n\times n$ matrix with the $n^2$ objects in random order and then take the rows as subsets (and clearly every split up into $n$ groups of $n$ is obtainable this way). There are $(n^2)!$ different orderings of the $n^2$ objects, but many lead to the same split-up: We can reorder each row in itself without changing the outcome, thus dividing the count by $n!$. As this applies to $n$ rows, we divide $n$ times, i.e., we divide by $n!^n$. Finally, the order of the rows is unimportant, i.e., we need to divide by yet another factor of $n!$. Final result: $$\frac{(n^2)!}{n!^{n+1}}. $$