“Distinguished element” in group theory I have asked that given a short list of triples of a set, a binary operation on the set, and a distinguished element, whether they forms a group. For example, the set is Z, the operation is addition and distinguished element is 1. In this context, what does distinguished element mean ? 

I know that the integer set with addition is a group with identity element 0, but if distinguished element means identity element, they don’t form a group.
I've seen these exercises before.  This is what your book is asking you:  given a trio of data as:


a set $G$
an operation $\ast$ on $G$
an element $x_0 \in G$ (your "distinguished element")


does $\ast$ give a group operation on $G$ with identity $x_0$?  Thus, for example, the answer for the data $(\mathbb{Z}, + , 1)$ is "no" but for the data $(\mathbb{Z}, + , 0)$ is "yes."
A relic of abstractionist parlance.  "Distinguished" just means "specific".  So, what they mean is that, in a group, the identity element is a specific element (with the required properties).  

Therefore, the answer to your question is: the triple $(\mathbb{Z}, +, 0)$ is a group, while the triple $(\mathbb{Z}, +, 1)$ is not.



Some more background to guard against future difficulties arising from the abstractionist approach--which some will find tangent and ranty--but, in the end, I will re-address your question in terms of what follows.

This misunderstanding (and many others) would not even arise if groups were introduced properly; i.e., as groups of transformations.

Definition:  A transformation of a set $X$ is a 1-to-1 correspondence from $X$ to $X$.

Corollary:  Composition of transformations of $X$ gives transformations of $X$, and this composition is associative.

This Corollary can be proved by direct inspection of composition.

Definition: A nonempty collection of transformations of $X$ is called a group of transformations of $X$ if this collection contains all compositions and inverses of its members.

Examples: i) $X$ = the Cartesian $xy$-coordinate plane, the transformations are the rotations of $X$ about the origin. ii) $X$ is a finite set, and the transformations are all permutations of $X$. iii) $X$ is any set, and the collection of transformations contains just one transformation: the one that maps every $x$ in $X$ to itself.  This transformation is called the identity transformation of $X$ will be denoted by $1_{X}$.

There is only one identity transformation: if there were two, multiply them, and the product equals each of them, hence they are the same.

Theorem:  If $G$ is a group of transformations of a set $X$, then $G$ contains the identity transformation $1_{X}$.

Proof:  If $g$ is a transformation in $G$, then so is $g^{-1}$ (by the latter definition), and therefore so is $g \circ g^{-1} = 1_{X}$.  Done.

If now we define an abstract group (the uninformative formalist way it's usually done: a set, a binary operation, a "distinguished" identity element, etc.), then:

Every abstract group is isomorphic to a group of transformations of some set.

(Why?  If $G_{abs}$ is an abstract group, then each $g \in G_{abs}$ gives us the transformation $x \mapsto gx$ of the set of the elements of $G_{abs}$.)

Therefore, groups of transformations really give us ALL the groups that exist. 

For finite groups, the version of this last boldfaced statement is Cayley's theorem.



To address your question in terms of groups of transformations:

Consider the shift of all integers by some constant integer (positive, negative, or zero). Such shifts are transformations of the set $\mathbb{Z}$. Among these, the identity transformation is the one that leaves all the integers in place; i.e., the shift by 0.  The shift by 1 is not the identity transformation.

That's why $(\mathbb{Z},+, 0)$ is a group, and $(\mathbb{Z},+,1)$ is not.