Why isn't $\mathbb Z_9 \cong \mathbb Z_3 \times \mathbb Z_3$ by the fundamental theorem of finite abelian groups?

The theorem does indeed hold: $\Bbb Z_9$ is a "trivial" direct product; it's cyclic, and just itself! Or, you can think of it as $\Bbb Z_9 \times \{0\}$, the product of itself with the trivial group (which is also cyclic!).

Just because

Every finite abelian group is the direct product of cyclic groups

that doesn't mean that every finite abelian group is the direct product of cyclic groups of prime order.

In general, one needs $\mathrm{gcd}(m,n) = 1$ for $\mathbb{Z}_{mn}$ to be isomorphic to $\mathbb{Z}_m \times \mathbb{Z}_n$.

More to the point: How many elements of order $9$ does $\Bbb Z_3\times\Bbb Z_3$ have?