Is this a correct way to think about specific examples of groups using the category theory definition? I'll say now, before anything else, that I probably don't know what I'm talking about. This is more me making a (hopefully) educated guess about a topic I'm not too familiar with. 
I recently started learning the basics of category theory - the absolute basics, as in what a category is as well as basic definitions. I'll admit I know practically no more than can be easily learned from a wikipedia article. 
I came across one category theoretical definition of a group - a single object category where every arrow is an isomorphism - and found it quite elegant. Intuitively it seemed more difficult to reconcile this definition with particular examples of groups, such as $(\mathbb{Z},+)$, than the usual set theoretical description. I've come up with one way of doing so specifically for $(\mathbb{Z},+)$, and I'm wondering if this kind of description is at all needed, or if it's wrong, or if there's a much simpler way of looking at it. 
I take the single object of $(\mathbb{Z},+)$ to be the set $\mathbb{Z}$ itself, and the arrows to be functions of the form $f:\mathbb{Z}\to\mathbb{Z}$, $f(x)=x+n$. I then simply write $f$ as $n$. Composition of arrows then is analogous to addition, as $m\circ n(x)=x+n+m$ could be written simply as $n+m$ (reversing the numbers is not necessary in this case, but it probably would be if addition weren't commutative). Clearly every integer exists as some such function, so this should contain all the information that the original set theoretical definition of $(\mathbb{Z},+)$ contained, anyway. 
Can most "concrete" examples of groups be represented in a similar way? I'd assume so, at least. Is this a valid way to think of these things?
 A: Let $(G, \circ)$ be a group. Let $\mathcal{C}$ be the subcategory of $Set$ consisting of the single object $G$, and $\mathcal{C}(G, G) := \{f_g: g \in G\} \subset Set(G, G)$, where $f_g$ is the function - i.e. morphism in $Set$ - satisfying $f_g(h) = h \circ g$ for all $h \in G$. Then we have $f_{g \circ g'} = f_g f_{g'}$, $f_e = 1_G$, and $f_g$ is an isomorphism for all $g$, with inverse $f_{g^{-1}}$ - that is, composition of functions makes $\mathcal{C}(G, G)$ into a group, and furthermore (easy check) the map $g \mapsto f_g$ is an isomorphism of groups between $G$ and $\mathcal{C}(G, G)$.
In short, the answer is yes. Any group can be represented in a similar way, even nonabelian ones.
A: You ask several questions, but I will only address the one in the title. No, this is in fact not the right way to think about groups as one-object all isos categories. While the details you give are correct, it is important to realise that the precise details of what the single object in the category is, is irrelevant. In fact, there is an equivalent definition of category in which there are no objects at all; the objects in a category are merely a convenience (albeit great convenience). 
