Constructing an isomorphism of group products

Why not adjust your try, to $C_1×\Bbb Z_{12}\cong \Bbb Z_3×\Bbb Z_4$, now that you know it was incorrect.

There are lots of examples like this, you can see after the other answer.

It is easier to find examples of numbers having two different factorizations, with both 'different from each other'.

For example take 4 distinct prime numbers $p,q,r,s$. Let $n$ be their product.

Now define $a=pq$ and $b=rs$. So $n=ab$. Now define $c=pr$ and $d= qs$. We still have $n=cd$. Clearly both are two distinct factorizations. $ab=cd$ with $a\neq c,d$ and $b\neq c,d$.

Now replace every integer above by cyclic groups of that order, and replace each factorization of two numbers, by direct product of corresponding groups, and we will get examples you are looking for.

Specific example: for the four prime numbers, $2,3,5,7$, their product is $210$. Using the notation $C_n$ for a cyclic group of order $n$, we see that $$ C_{210}\cong C_6\times C_{35}\cong C_{10}\times C_{21}$$

We are using the fact that for two finite cyclic groups of coprime orders, their direct product is again a cyclic group.

All this is elementary number theory. Examples involving non-abelian groups would be a bit more challenging and would be real group theory.