Nonisomorphic Groups of order $100$

You are wrong, $\mathbb Z_{100}$ is an option. But it appears twice in your list: Note that $4$ and $25$ do not share any prime factors, so by the Chinese Remainder Theorem, $$\mathbb Z_4 \times \mathbb Z_{25} \cong \mathbb Z_{100}.$$ As pointed out in the comments, an elementary way to see this is as follows. The element $(1, 1) \in \mathbb Z_4 \times \mathbb Z_{25}$ has order $100$, because if $n$ is a multiple of $4$ and a multiple of $25$, then it is a multiple of $\text{lcm}(4, 25) = 100$. So $(n, n)$ is zero in $\mathbb Z_4 \times \mathbb Z_{25}$ for no $n$ below $100$. Thus, there are at least $100$ elements in this group.

$\Bbb Z_{100}$ is isomorphic to $\Bbb Z_4\times\Bbb Z_{25}$.

Personally, I prefer to let each factor have prime power order when solving an exercise like this. That way it's much easier to check both that I have all options and that I have no repeats. In this case it would be $$ \Bbb Z_4\times\Bbb Z_{25}\\ \Bbb Z_2\times\Bbb Z_2\times\Bbb Z_{25}\\ \Bbb Z_4\times\Bbb Z_5\times\Bbb Z_5\\ \Bbb Z_2\times\Bbb Z_{2}\times \Bbb Z_5\times\Bbb Z_5 $$

Since $4$ and $25$ are relatively prime, $\mathbb Z_4\times \mathbb Z_{25}\simeq \mathbb Z_{100}$, which is why you have an extra one on your list. Of course $\mathbb Z_{100}$ is a valid example, as it is an abelian group of order $100$.

The way to ensure you don't have extras is to make sure all of the factors have prime power order, so while $\mathbb Z_{100}$ is correct, it would not show up on your list as such, but rather as $\mathbb Z_4\times\mathbb Z_{25}$.