In additive abelian group the word ‘additive’ refers to the symbol used for the operation $({+})$ and, in principle, it has nothing to do with the group being abelian. It's true that in most cases the additive notation is used for abelian groups (or, more generally, for commutative operations), but this is not universal.
For instance, the two operations on near-rings are usually denoted by addition and multiplication, but addition is not required to be commutative although a near-ring must be a group with respect to addition (see http://en.wikipedia.org/wiki/Near-ring).
You need to show that for any two elements $a, b$ in $G$, $a*b = b*a$.
$$a*b = \dfrac{\color{blue}{a\cdot b}}2 = \dfrac{\color{blue}{b\cdot a}}2 = ba$$
Because "normal" multiplication on the real numbers is commutative, we know $\color{blue}{a\cdot b = b\cdot a}$. Hence, we have commutativity.
Now you need to establish that $G$ is in fact, a group:
Associative? Check whether, given any $3$ elements $a, b, c$ in $G$, is it true that $$a*(b*c) = (a*b)*c\;\;?$$
Existence of an Identity element $e$ such that for all $a \in G$, $a*e = e*a = a$? Let's check out whether $2$ meets this criteria: $$a*2 = \dfrac {a \cdot 2}2 =\color{blue} a = \dfrac{2\cdot a}2 = 2*a$$
Closure under inverses?: For every element $a\in G$, does there exist an element $a^{-1}$ such that $aa^{-1} = a^{-1}a = 2= e$? Let's check out whether $\frac 4a = a^{-1}$: $$a*\frac 4a= \frac {a\cdot \frac 4a} 2 = \frac 42 = \color{blue}2 = \frac{\frac 4a \cdot a}2 = \frac 4a*a$$
Once you confirm that associativity holds, we'll see that $[G, *]$ is a commutative group, i.e. and abelian group.
Best Answer
$(G,\cdot)$ denotes the ordered pair with first entry the underlying set $G$ of the group and second entry the law of composition $\cdot$ of the group. The ordered pair notation is very common in other parts of mathematics, for instance for metric spaces $(M,d)$ or topological spaces $(X,\tau)$ and so on (the general pattern is $(\mathrm{set},\mathrm{structure \ on \ the \ set})$; it is useful because one does not have to define the notion of equality for groups, metric spaces and topological spaces and so on separately as it can be shown that two ordered pairs $(x,y)$ and $(x',y')$ are equal as sets iff $x=x'$ and $y=y'$. Hence two groups are equal iff their underlying set and their law of composition are equal. It is however very common to denote the group $(G,\cdot)$ simply by $G$ (similarily for metric and topological spaces) if the law of composition (metric, topology) is understood. Also a group is by definition an ordered pair $(G,\cdot)$ with $G$ a set and $\cdot$ a law of composition on $G$ subject to the group axioms (some people say that a group is a set $G$ "together with a law of composition on $G$" by which they simply mean that $(G,\cdot)$ is an ordered pair).
Additive notation refers to denoting the law of composition by $+$ (multiplicative notation $\cdot$), the unit element by $0$ (multiplicative notation $1$) and the inverse of $x$ by $-x$ (multiplicative notation $x^{-1}$). As others have pointed out, this is by convention often done when the law of composition is commutative.