TRUE/FALSE :Any two abelian group of order 8 must be isomorphic
SOLUTION: True
The problem of finding all abelian groups of order 8 is impossible to solve, because there are infinitely many possibilities. But if we ask for a list of abelian groups of order 8 that comes with a guarantee that any possible abelian group of order 8 must be isomorphic to one of the groups on the list Z8, Z4 × Z2, Z2 × Z2 × Z2.
Your question/answer and what you write at the bottom contradict one another. There are $3$ abelian groups of order $8$, up to isomorphism. And you've listed them:
$$\mathbb Z_8,\;\;\mathbb Z_4\times \mathbb Z_2,\, \text{ or }\,\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2.$$
Among those three groups, no two of them is isomorphic. So it is false any two abelian groups of order $8$ are isomporphic.
$$\begin{align}\mathbb Z_8&\not\cong \mathbb Z_4\times \mathbb Z_2& \mathbb Z_4\times \mathbb Z_2 &\not\cong \mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2& \mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2&\not\cong \mathbb Z_8 \end{align}$$
It is true that all abelian groups of order $8$ are isomorphic to one of the three non-isomorphic groups: $$\mathbb Z_8,\;\;\mathbb Z_4\times \mathbb Z_2,\, \text{ or }\,\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2.$$