Yes, you're right.
Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?
In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form
$K^* \cong \newcommand{\Z}{\mathbb{Z}}$
$\Z/n\Z \oplus \bigoplus_{i=1}^{\infty} \Z.$
Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.
The appearance of the term "Betti number" in your theorem should be seen as a definition. The Betti number of a finitely generated abelian group $G$ is the number of $\mathbb{Z}$ factors appearing in the decomposition claimed by your theorem, or differently stated, the rank of $G$ as a $\mathbb{Z}$-module.
You can also understand it as a sort of analogue to the dimension of a vector space. Indeed, tensoring $G$ with $\mathbb{Q}$ will convert it into a $\mathbb{Q}$-vector space whose dimension is exactly the rank/Betti number of $G$.
That is, for example, if $G=\mathbb{Z}^n$, then $n$ is the Betti number.
Another example is $G=\mathbb{Z}_{(p)}$. There is no $\mathbb{Z}$ factor, so the Betti number is $0$.
Now let us look at your two statements:
The first one is false: Take $G=\mathbb{Z}$ and $H=\mathbb{Z}\times \mathbb{Z}_{(2)}$. They both have Betti number $1$, but they are clearly not isomorphic.
The second one is true: if the Betti number was positive, then the group would have a subgroup isomorphic to $\mathbb{Z}$, in particular it would have infinitely many elements.
Best Answer
See this question and the answer by Jim Belk. The group is one of the "relatives" of the R.Thompson group $T$. This one is obtained by lifting Thompson's group $T$ through the covering map from the line to the circle. It is a natural finitely generated (even finitely presented) group, and it has been considered many times before by Ghys, Sergiescu and others.