Nonisomorphic groups of order 12.

The groups you've found so far:

$$ A_4, \;\mathbb Z_{12}, \mathbb Z_6\times \mathbb Z_2$$ are, indeed, non-isomorphic. Why?

By the Fundamental Theorem of Finitely Generated Abelian groups, we know $\mathbb Z_{12}$ and $\mathbb Z_6\times \mathbb Z_2 \cong \mathbb Z_2\times \mathbb Z_6$ are abelian and further, that $\mathbb Z_{12}\cong \mathbb Z_3 \times \mathbb Z_4$ and is cyclic, and not isomorphic to the abelian group $\mathbb Z_{2}\times \mathbb Z_6$.

$\mathbb Z_{12} = \mathbb Z_{2\times 6} \not\cong \mathbb Z_2\times \mathbb Z_6$ because $\gcd(2, 6) = 2\neq 1$. Indeed, $\mathbb Z_{12}$ is cyclic, but $\mathbb Z_2\times \mathbb Z_6$ is not.
Neither of these two non-isomorphic abelian groups is isomorphic to $A_4$, since $A_4$ is not abelian.

Finally, for a fourth group of order $12$ which is not isomporphic to any of the above three groups, we have $\;\mathbb Z_2\times S_3$. This group is not abelian, and so not isomorphic to $\mathbb Z_{12},$ nor to $\mathbb Z_2\times \mathbb Z_6$. All that's left for you to justify is the fact that $A_4\not\cong \mathbb Z_2 \times \mathbb S_3$.

Yes, external direct product is need. You can get $A_4$ and $Z_2\times S_3$ are not isomorphic. And you can get all groups in abelian case easily.

You can use the external direct product for this problem.

There are actually five non-isomorphic groups of order $12$, but you need to find four. Here are few catches to determine the specific groups:

Apply the fundamental theorem of finitely-generated abelian groups. Be careful of which groups are isomorphic.
Try to find other groups by constructing homomorphism and applying stabilizer theorem.