I just read that if $G$ is an abelian group with subgroup $A$, then we could not always find a subgroup $B$ such that $G = A \oplus B$.
I tried to come up with an example, let $G = \mathbb Z^{\mathbb N}$ and $A := \langle (1,1,1,1,1,\ldots ) \rangle$, suppose $G = A \oplus B$ and consider $(0,-1,-1,-1,\ldots) = (-1,-1,-1,-1,\ldots) + (1,0,0,0,\ldots)$, which is a unique decomposition with an element from $A$, hence we must have $(1,0,0,0,\ldots) \in B$, by considering $(-1,0,-1,-1,\ldots)$ we find $(0,1,0,0,\ldots ) \in B$ and so on, and as these elements generate $G$ this implies $G = B$, a contradiction to $G = A \oplus B$; hence we could not have such an direct sum decomposition.
Is my above reasoning correct? Do you know any other examples?
Your argument is not conclusive, because the elements you list do not generate $\mathbb{Z}^{\mathbb{N}}$, but only $\mathbb{Z}^{(\mathbb{N})}$ (the subgroup of sequences with only a finite number of nonzero terms).
A much simpler example is $G=\mathbb{Z}$ and $A=2\mathbb{Z}$.
There's no $B$, because two nonzero subgroups of $\mathbb{Z}$ always have nonzero intersection.
Groups with the property that no nonzero subgroup has a complement are indecomposable and there are many: $\mathbb{Z}/p^n\mathbb{Z}$ with $p$ a prime, $\mathbb{Z}$, $\mathbb{Q}$ are basic examples.
By the way, $\mathbb{Z}^{\mathbb{N}}$ is decomposable: take $A=\langle(1,0,0,\dotsc)\rangle$ and $B=\{(0,a_1,a_2,\dotsc):a_i\in\mathbb{Z}\}$.