That the morphism induced by the inclusion $\mathbb{Z}[[x]]\to\mathbb{Q}[[x]]$ is not an isomorphism can also be derived from group theoretical properties.
As abelian groups, $\mathbb{Z}[[x]]$ and $\mathbb{Q}[[x]]$ are just direct products of copies of $\mathbb{Z}$ and $\mathbb{Q}$. Consider the exact sequence
$$
0\to \mathbb{Z}^{\mathbb{N}}\to
\mathbb{Q}^{\mathbb{N}} \to
(\mathbb{Q}/\mathbb{Z})^{\mathbb{N}}
\to0
$$
Tensoring with $\mathbb{Q}$ is right exact and we get the exact sequence
$$
\mathbb{Z}^{\mathbb{N}}\otimes\mathbb{Q}\to
\mathbb{Q}^{\mathbb{N}}\otimes\mathbb{Q} \to
(\mathbb{Q}/\mathbb{Z})^{\mathbb{N}}\otimes\mathbb{Q}
\to0
$$
but the group $(\mathbb{Q}/\mathbb{Z})^{\mathbb{N}}$ is not torsion, so
$(\mathbb{Q}/\mathbb{Z})^{\mathbb{N}}\otimes\mathbb{Q}\ne0$, which means
$\mathbb{Z}^{\mathbb{N}}\otimes\mathbb{Q}\to \mathbb{Q}^{\mathbb{N}}\otimes\mathbb{Q}$ is not surjective.
(Note that the map $\mathbb{Z}^{\mathbb{N}}\otimes\mathbb{Q}\to
\mathbb{Q}^{\mathbb{N}}\otimes\mathbb{Q}$ is injective because $\mathbb{Q}$ is flat, but it's irrelevant for the problem at hand.)
However, $\mathbb{Z}^{\mathbb{N}}\otimes\mathbb{Q}$ and $\mathbb{Q}^{\mathbb{N}}\otimes\mathbb{Q}$ are isomorphic (through a different map). Indeed they are both torsion free divisible groups (that is, $\mathbb{Q}$-vector spaces) with the same cardinality $2^{\aleph_0}$, so they have the same dimension $2^{\aleph_0}$ (assuming choice, of course). However, as shown in another answer, this group isomorphism cannot be a ring homomorphism.