The main point is that a group scheme does not have enough points :-)
For example, given any number field $K$ (=finite-dimensional extension of $\mathbb Q$), Mazur-Rubin have recently proved that there exists an elliptic curve $E$ defined over $K$ such that $E(K)$ (=points with coordinates in $K$) is the trivial group with one element: this is absurdly insufficient for studying $E$.
More importantly the possibility to vary the field (or even scheme) and look at points of the group-scheme with coordinates in that field, i.e. the morphism point of view, is incredibly important in arithmetic geometry.
For example, if $E$ is an elliptic curve defined over $\mathbb F_p$ it has a zeta function $Z(E)= \operatorname {exp} (\sum\operatorname {card}E(\mathbb F_{p^n})\cdot\frac{T^n}{n}) $
which is, amazingly, a rational function in $T$ and has been the prototype of the arithmetic study of group-schemes.
That said, over an algebraically closed field $k$ , the classical study of a group-scheme just as a group $G(k)$ is quite interesting and fruitful: it is called the geometric study of $G$, as opposed to the arithmetic study evoked above.
Edit: points of a scheme
Given a scheme $X$ , a point with values in a another scheme $T$ is a morphism $T\to X$. This looks rather opaque (especially if you also consider that all schemes are over another scheme $S$ !) so let me try and show that this notion is not so unreasonable as it looks.
Consider the field (or even ring) $k$, the algebra $A=k[T_1,...,T_n]/I$ ( where $I\subset k[T_1,...,T_n]$ is an ideal) and the affine $k$-scheme $X=Spec(A)$.
Now, what is a $T$-point of $X$ if you take for $T$ the affine scheme $T=Spec(R)$ corresponding to some $k$-algebra $R$ ?
Answer: it just corresponds to $V(R)\subset R^n$, the set of solutions $(r_1,...,r_n)\in R^n$ of the system of equations $P(r_1,...,r_n)=0 \quad \text{for all} \;P(T_1,...,T_n)\in I$ which is the naïve interpretation of "points of $X$ with values in $R$".
And this is easy: a $k$-morphism $Spec(R)\to X=Spec(A)$ corresponds to a morphism of $k$-algebras $\phi: A=k[T_1,...,T_n]/I\to R$ to which you just associate the naïve $n$-tuple $(r_1,...,r_n)=(\phi(\bar T_1),...,\phi(\bar T_1))\in V(R)$