Structure ring of constant group scheme.

For the finite abelian group $G$, the group scheme $G_{\mathrm{Spec}\,{\Bbb Z}} = {\mathrm{Spec}}\,{\cal O}_G$ over ${\mathrm{Spec}}\,{\Bbb Z}$ is defined as follows$\colon$ $$ {\cal O}_G = {\Bbb Z}e_0 \oplus {\Bbb Z}e_1 \cdots \oplus {\Bbb Z}e_{g} $$ with $$ e_g^2 = e_g, e_g e_{g'} = \delta_{g,g'}, m(e_g) \colon= \!\!\!\underset{g',g''| g' + g'' = g}{\Sigma} e_{g'} \otimes e_{g'}, \iota(e_g) \colon= e_{-g}, \epsilon(e_g) = \delta_{0,g}. $$

Q. Does it hols that $\pi \colon {\cal O}_G \cong k[G]$, where $k[G] = k[u_g ; g \in G]$? How can I make an explicit isomorphism $\pi$?

The question is ``isomorphic as what?''. It is obvious that $\mathcal{O}_G \cong \mathbb{Z}[G]$ as $\mathbb{Z}$-modules because they are $\mathbb{Z}$-modules of the same rank.

However, we want to compare these objects with their multiplication and comultiplication structures. You should be able to convince yourself that $\mathcal{O}_G$ and $\mathbb{Z}[G]$ cannot be isomorphic as rings.

For example let $G =\{ e, \sigma \}$ with $\sigma^2 = e$. Then both $\mathbb{Z}$-modules have a basis $e,\sigma$.

In $\mathcal{O}_G$ we have $(a e + b \sigma)^2 =a^2 e + b^2 \sigma$ so $e$ and $\sigma$ are idempotents.

In $\mathbb{Z}[G]$ we have $(a e + b \sigma)^2 = (a^2 + b^2) e + 2 ab \sigma$ so if $(a e + b \sigma)$ is idempotent then $a = a^2 + b^2$ and $b = 2 ab$ so either $b = 0$ and $a = 1$ or $a = 0$ and $b = 0$. Therefore there are only trvial idempotents in this ring.

However, these objects have more structure, they are Hopf algebras meaning they also carry a compatible comultiplication structure (you wrote down the comultiplication $m$ for $\mathcal{O}_G$ and for $\mathbb{Z}[G]$ this is given by $u_g \mapsto u_g \otimes u_g$). Then it turns out that $\mathbb{Z}[G]$ and $\mathcal{O}_G$ are dual Hopf algebras, $\mathbb{Z}[G] = (\mathcal{O}_G)^*$. You can think of this operation as ``swapping multiplication and comultiplication'' if that makes it easier to visualize what is going on. The isomorphism takes $u_g \mapsto (e_g \mapsto 1)$ which makes sense thinking of $\mathbb{Z}[G]$ as functions on G

We can promote this to a duality of finite locally free group schemes. Saying that on the functor of points, $$ G^\vee(T) = \mathrm{Hom}_{\text{Grp}_T}(G_T, \mathbb{G}_m) $$ Then we see that $\mathrm{Spec}(\mathbb{Z}[G])$ is the dual group to $G_{\mathrm{Spec}(\mathbb{Z})}$

For example, if $G = \mathbb{Z}/n \mathbb{Z}$ then $\mathbb{Z}[G] = \mathbb{Z}[x]/(x^n - 1)$ so,

$$ G_{\mathrm{Spec}(\mathbb{Z})}^\vee = \mathrm{Spec}(\mathbb{Z}[G]) = \mathrm{Spec}(\mathbb{Z}[x]/(x^n - 1)) = \mu_n $$ is the group of $n^{\text{th}}$-roots of unity.