The fact that ordering does not matter within a group is already taken care of by the binomial coefficients. The additional $2!$ and $3!$ you see in the answer are taking care of the fact that the order in which the groups themselves were chosen also does not matter.
For example, if your two-person groups are $\{A, B\}$, $\{C, D\}$, and $\{E, F\}$, then the following arrangements are all the same:
$\{A, B\}$, $\{C, D\}$, $\{E, F\}$
$\{A, B\}$, $\{E, F\}$, $\{C, D\}$
$\{C, D\}$, $\{A, B\}$, $\{E, F\}$
$\{C, D\}$, $\{E, F\}$, $\{A, B\}$
$\{E, F\}$, $\{A, B\}$, $\{C, D\}$
$\{E, F\}$, $\{C, D\}$, $\{A, B\}$
Notice there are $3!$ such arrangements. When you just multiply your binomial coefficients together, however, these all get counted as distinct. Dividing by $3!$ collapses these all into a single arrangement.
To give another example with a better selection of numbers, suppose you want to arrange 6 people into three groups of two each. This would be given by
$$
\frac{\binom{6}{2} \binom{4}{2} \binom{2}{2}}{3!}.
$$
Again, the $3!$ is coming from the number of groups, not their size.
Best Answer
It sounds like the reasoning behind your argument is as follows: since each group needs at least one person, use ${n\choose 2}$ to choose those two people, sort them into groups in one of $2$ ways, and then use $2^{n-2}$ to decide where the rest of the people should go. Here's the problem with that argument: the two people you chose at the beginning are not special, but are made special by your count. My faulty argument said to "choose those two people", but those two people chosen did not have to be anyone in particular, as long as we they were put into different groups.
For example, say $n=4$. One way to sort the groups would be to put $A$ and $B$ in group $1$, and $C$ and $D$ in group $2$. Your counting method would count this arrangement $4$ times. Your ${n\choose 2}$ could pick $A$ and $C$, then your method would put them in groups $1$ and $2$ respectively and distribute the rest of the people. However, it could also pick $A$ and $D$ and do the same, as well as for $B$ and $C$, and $B$ and $D$. This counts the same group multiple times because it treats the two people you chose at the beginning as being special, when they're not. That said, your argument does solve a similar, but different counting problem: given $n$ people, count the number of ways to divide them into two distinct groups, where each group has a leader.
To get the correct answer, ask yourself the following: how many ways are there to divide $n$ people into two groups? How many ways of doing that fail to satisfy the condition that each group is nonempty?