Whether such an algebra has to be the Group algebra Let $\mathbb C$ be the field of the complex numbers, $\mathbb Q$ the field of the rational numbers.
Let $G$ be an additive subgroup of $\mathbb Q$.
$R$ is an commutative algebra over $\mathbb C$, which is a domain and $G$-graded with all the $G$-graded spaces are 1-dimensional.
Is it true that we have $R \cong {\mathbb C}[G]$(the group algebra over $G$)?
 A: Yes, such associative, not necessarily commutative algebras are classified by $H^2(G, {\mathbb C}^\ast)$. Indeed, try to define new multiplication on the group algebra by $g\cdot h = \alpha (g,h) gh$. Then $\alpha$ must be a cocyle, and isomorphic algebras correspond to cohomologous cocycles.
Finally,  $H^2(G, {\mathbb C}^\ast)= Hom(\Lambda^2 G, {\mathbb C}^\ast)= 0$ since any 2 elements are linearly dependent.  See here for the calculation of $H^2$.
A: Yes, it's isomorphic to the group algebra. We can do even a bit better: 

Let $G$ be a torsion-free abelian group those finitely generated subgroups have rank one and let $k$ be an algebraically closed field. If $R$ is an associative zero-divisor-free $G$-graded $k$-algebra with $\dim_k R_g=1$ for all $g \in G$, then $R \cong k[G]$. 

The proof is along Bugs' lines. Denote the product in $R$ by $\circ$ and write $g \circ h = a(g,h)gh$, $a(g,h) \in k$. Since $R$ is zero-divisor-free this defines a map $a: G \times G \to k^\times$. By associativity, $a$ satisfies for all $f,g,h \in G$: 
$$a(f,g)\cdot a(fg,h)=a(g,h)\cdot a(f,gh).$$
Futhermore, since the neutral element $e \in G$ is also the identity of $R$, we have 
$$a(f,e) = 1 = a(e,f).$$
These two equations just say that $a$ is a normalized cocycle, if we consider $k^\times$ as a trivial $G$-module (cf. (3.4),(3.9) in chap. IV of Brown: Cohomology of Groups). So $a$ represents an element in $H^2(G,k^\times)$. 
Assume $H^2(G,k^\times)=0$. Then $a$ is a coboundary, i.e. there is a map $b: G \to k^\times$ such that $$a(g,h)=b(g)b(h)b(gh)^{-1}.$$ Define $\varphi: k[G] \to R,\; g \mapsto b(g)^{-1}g$. Obviously, $\varphi$ is an isomorphism of $k$-vector spaces and because of 
$$\varphi(g) \circ \varphi(h)=b(g)^{-1}b(h)^{-1}(g \circ h) = b(g)^{-1}b(h)^{-1}a(g,h)gh=b(gh)^{-1}gh=\varphi(gh)$$
it's an isomorphism of graded $k$-algebras. 
Hence it suffices to show $H^2(G,k^\times)=0$. By universal coefficients (Hilton-Stammbach: A course in homological algebra, Theorem 15.1) there is a short exact sequence 
$$0 \to Ext_{\mathbb Z}^1(H_1(G),k^\times) \to H^2(G,k^\times) \to Hom(H_2(G),k^\times) \to 0$$
where $H_n(G) := H_n(G,\mathbb Z)$. As an abelian group, $G$ is the direct limit of its f.g. subgroups $H$ and because homology commutes with direct limits, $H_2(G)$ is the direct limit of $H_2(H)$. But by assumption $H \cong \mathbb Z$, whence $H_2(H)=0$.
Since $k$ is algebraically closed, the multiplicative group $k^\times$ is a divisible $\mathbb Z$-module and thus injective (Brown, III, 4.2). Hence $Ext_{\mathbb Z}^1(H_1(G),k^\times)=0$ and the short exact sequence implies $H^2(G,k^\times)=0$. q.e.d.
Remark: As the proof shows, the condition that $R$ is zero-divisor-free can be lowered to the assumption $g\circ h \neq 0$. 
