Consider the two abelian groups $G_r=\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}$ (with $r$ copies of $\mathbb{Z}$, $r\ge 1$) and similarly $G_s=\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}$, $s\ge 1$.
If $f:G_r\rightarrow G_s$ is a group isomorphism, then, $f$ should carry the subgroup $2G_r=\{x+x\,\,|\,\, x\in G_r\}$ into $2G_s = \{ y+y\,\,|\,\, y\in G_s\}$.
Hence, we can conclude that $G_r/2G_r \cong G_s/2G_s$.
Now, the last two groups are finite groups of order $2^r$ and $2^s$ respectively, so we can conclude that $r=s$.
Question: In some standard books, such as The Theory of Groups by A. G. Kurosh, or Basic Algebra by Anthony Knapp, I saw that to define rank of a finitely generated free abelian group, the authors deal with generating sets, relations, and then proving $r\ge s$ and $r\le s$, to conclude that $r=s$. So, I was wondering why the above elementary arguments are not used (even in the very modern text by Knapp).
The proof with generating sets and linear independence is “more general”, because the technique can be employed in other contexts.
Your approach is fine, however, and can be even “simplified”. If you have any field $F$, then you have a bimodule ${}_{\mathbb{Z}}F_F$ and so, given any homomorphism $f\colon G_1\to G_2$ of abelian groups, we get a homomorphism of $F$-vector spaces $$ f\otimes_{\mathbb{Z}}F\colon G_1\otimes_{\mathbb{Z}}F\to G_2\otimes_{\mathbb{Z}}F $$ If $f$ is an isomorphism, then also $f\otimes_{\mathbb{Z}}F$ is; if $G_1$ is finitely generated, then $G_1\otimes_{\mathbb{Z}}F$ is finite dimensional. If $G_1=\mathbb{Z}^r$, then $G_1\otimes_{\mathbb{Z}}F\cong F^r$.
With $F=\mathbb{Z}/p\mathbb{Z}$, you get, for instance, the concept of $p$-rank, which is an invariant as well as the standard rank (obtained with $F=\mathbb{Q}$).