Are these isomorphic $\mathbb{Z}_{2}\times\mathbb{Z}_{3}$ and $\mathbb{Z}_{9}^{*}$ Is  $\mathbb{Z}_{2}\times\mathbb{Z}_{3}$ isomorphic to $\mathbb{Z}_{9}^{*}$
both have orders 6
both have elements with orders 1,2,3,6 (1 element of order 1, 2 elements of order 3, 1 element of order 2 and 2 elements of order 6) 
Both are cyclic thus Abelian
Can I assume then that they are isomorphic? Or a specific isomorphism must be constructed?

Yes, the two groups are isomorphic. And you are almost there in proving this.
Note that $\mathbb Z_2\times \mathbb Z_3 = \mathbb Z_6$, since $\gcd(2, 3) = 1$.
And since the order of $\mathbb Z^*_9 = 6$ and is cyclic, we know that $\mathbb Z^*_9 \cong \mathbb Z_6$. 
There is no need to construct an explicit isomorphism here, though you need to (implicitly or explicitly) invoke the following standard fact about finite cyclic groups:

Every finite cyclic group of order $n$ is isomorphic to $\mathbb Z_n$, the group of integers under addition, modulo $n$.

