Is the definition of the binary operation of a group intrinsic to the set elements? I have a nagging, basic question as to the boundaries between elements in the set and composition rule (or operation) in the definition of a group:

In mathematics, a group is an algebraic structure consisting of a set of elements equipped with an operation that combines any two elements to form a third element...

For instance, in a finite cyclic group of order $n=4,$ i.e. $\mathbb Z_4=\{0,1,2,3\}$ the element $1$ moves all elements of the set by $1,$ while the element $2$ moves them by $2,$ etc.
Now in the composition table, as an example, $1\;\circ\;3$ will move the elements first by $1,$ and later by $3,$ resulting in no move $(\mathbb Z/ 4\mathbb Z).$
The point being that the "effect" (I would like to say "action", but the word is taken to signify the action of a group on a set) of the binary operation is already somehow within the actual elements.
In other words, the elements of the set are not inert or passive. They don't seem to be waiting for the definition of the operation - they carry with them a generating "action", which seems to overlap or preempt the role of the binary operation. At least in this example, they don't seem to be in need for an explicit definition of the composition rule.

Up to what point are the elements "active", and already intrinsically containing the idea of the binary operation? Is the composition implicit in the set elements, relegating the composition table to a formalism? Or does it depend on the specific type of group?

Another example: What else needs defining in the dihedral group $D_3$ when the elements are "rotation" and "reflection", that is $S=\{e,r,r^2,rf,r^2f\}?$
Please be lenient with terminological slights of hand to convey the conceptual problem. Perhaps fixing the terminology can clarify the issue.
 A: Good that you like to think "philosophically" about what things are.  I think what behind your observation is that groups often and most naturally appear as symmetry groups of something. In that case what a group element "is" is described by what it does, and the description  also contains the instructions for the group operation, since group multiplication is composition of symmetries.
Of course, you should not try to make too much out of this idea -- it is just one way of looking at things.  The same group can be realized as symmetries of different things, and different realizations will give you a different view of what the group multiplication means. Also a group can be described in one way or another that doesn't lend itself to seeing the group as a symmetry group -- in that case the names of the group elements will tell you nothing about the group operation.  
A: Pretty much the opposite is true. You could take that same set, $\{0,1,2,3\}$ and make the following multiplication table:
\begin{array}{l|llll}
  &0&1&2&3\\
\hline
0&2&3&1&0\\
1&3&2&0&1\\
2&1&0&3&2\\
3&0&1&2&3
\end{array}
I've simply flipped the roles of $3$ and $0$. The binary operation is really all that matters.
Given a binary operation $*:G\times G\to G$, an element $x$ does induce a function $L_x:G\to G$ by defining
$$L_x(y)=x*y$$
This is the associated left multiplication map. Similarly there is one for right multiplication. Perhaps the existence of these maps is what you're observing.
A: $\mathbb{Z_4} = \{0,1,2,3\}$ is not a group: it is a set. Of course this is a silly distinction that nobody actually makes in practice, but I think it is an important one for you, since you seem to be thinking:
"If a set contains elements for which there is some 'natural' binary operation that satisfies the group axioms, it is a group." 
To illustrate: Does $\{Rock, Paper, Scissors\}$ form a group? Well, of course there's a very natural operation, one that you might say "preempts" a binary operation, or makes the elements "active". We have "$vs$", which gives: $$Rock \space vs \space Paper = Paper$$ $$Scissors\space vs\space Rock = Rock$$ $$Paper \space vs \space Scissors = Scissors$$ and so on. You might say that it doesn't form a group, maybe because there's no identity (i.e. no element that loses to everything). 
But what if we defined some other operation, say "$+$", so that our set acts like $\mathbb{Z}_3$ with addition modulo 3 (which obviously obeys group axioms)? Is it now suddenly a group? 
On the other hand, take $S = \{e,r,r^2,rf,r^2f\}$ -- is this a group? According to your view, seemingly yes... but we know we can easily take some unruly binary operation $\ast$ that doesn't satisfy the axioms. How is this operation any less 'natural' than function composition? 
The point is this: just because the elements of a set are involved in a previously-defined binary operation, it doesn't mean it's the "canonical" binary operation for which to base the group on. When we say "$\mathbb{Z}_4$ is a group", this is literally just shorthand for "$\mathbb{Z}_4$ taken with the typical mod 4 addition operation forms a group" or "$(\mathbb{Z}_4, +_4)$ is a group, where $+_4$ is the obvious thing". 
Basically: It literally makes no sense to say something is a group because of the nature of the elements it contains, with no mention of the particular operation you're choosing. 
