What does the notation $\mathbb{Z} e_n$ mean? What does the notation in the direct sum $\underset{n \in \delta}{\oplus} \mathbb{Z} e_n$ means? Why is there an $e_n$, isn't it an element of $\lbrace -1,1 \rbrace$ and does not changes anything?
The $\delta$ is the cardinality and the context is the direct sum of rational groups.
paper
Pure subgroups of completely decomposable groups - an algorithmic approach. By Daniel Herden and Lutz Strüngmann, page 5
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.547.104&rep=rep1&type=pdf
 A: Whenever you have some set $M$ you can consider the free abelian group $\mathbb{Z}^{\oplus M} = \bigoplus_{m \in M} \mathbb{Z}m$. That means you take one copy of $\mathbb{Z}$ for every element $m \in M$ and treat that element like $1$ (or $-1$ if you wish). So the free abelian group with basis $\lbrace x,y,z \rbrace$ is given by formal expressions of the form $ax + by + cz$, where $a,b,c \in \mathbb{Z}$. For this construction it does not matter at all what type of object the elements in $M$ are.
I did not check what your $e_n$ are, but as you can see they do not have to be $1$ or $-1$. It still works if they are noodles, cars or whatever you want. It will still be useful to figure out what they are as you are using them as a basis here. 
Maybe the author really just picks $e_n \in \lbrace \pm 1 \rbrace$ and uses that notation to have $(e_1,0,\dots,0)$ etc as basis vectors later on without choosing between $\pm 1$.
A: I assume the $e_n$ are identities of a family of algebraic objects.
Omitting the $e_n$ gives you an isomorphic object, but including the $e_n$ makes clear that you work with that object as subobject of some other object.
