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let $G$ be a group and let $\mathcal{A}[G]$ be the group algebra of $G$.

I want to show the statement from the book "$G$ is a subgroup of the group of invertible elements of $\mathcal{A}[G]$."

So first I want to determine which elements are invertible:

$$(\sum_{g\in G} x(g)\delta_g)*(\sum_{h\in G}y(h)\delta_h)=\sum_{g,h\in G}x(g)y(h)\delta_{gh}=\delta_e$$

So all terms where $gh\ne e$ vanish $\implies$ $$=\sum_{gh=e} x(g)y(h)\delta_e$$ $$\implies \sum_{gh=e}x(g)y(h)=1$$ $$\implies \sum_{g\in G}x(g)y(g^{-1})=1$$

I am not sure how to get from here to the statement. For one thing, I can't see in what way $G$ can be a subgroup of $\mathcal{A}[G]$. I imagine they mean an isomorphic copy of $G$ or something?

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    If you think of $\mathcal{A}[G]$ as formal linear combinations of elements of $G$, that already makes $G$ a subset of $\mathcal{A}[G],$ and the multiplication operation in $\mathcal{A}[G]$ restricted to the subset $G$ is simply the usual group operation. – 2017-02-16

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The point is that you prove that $G$, interpreted as a subset of $\mathcal{A}[G]$, is contained in the set of invertible elements.

Don't try to prove the other way round (as you did) since you won't be lucky (see e.g. Kaplansky'S conjectures). (as you wrote group algebra I assume you mean that $\mathcal{A}$ is a field) It is simply not true that the invertible elements of $\mathcal{A}[G]$ "look" exactly like the elements of $G$. Think of "multiples" of elements of $G$ for example (again see Kaplansky's conjectures).

So write down an element of $G$ in its "shape" as an element of $\mathcal{A}[G]$ and try to find its inverse in $\mathcal{A}[G]$. I guess you will see that this is not a hard task.

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    I think if $\mathcal{A}=\mathbb{F}_2$, and if $G$ is abelian and torsion-free, then $(\mathcal{A}[G])^{\times}\simeq G$ (I saw it in a paper, I can guve references if necessary); but it won't be the case more generally – 2017-02-16