Let $R$ be a ring and $\mathop{\mathrm{Spec}}A$ be a finite flat commutative group scheme over $\mathop{\mathrm{Spec}}R$ so that the theory of Cartier duality applies.
Denote $s:R\to A,\ m:A\to A\otimes_R A,\ i:A\to A,\ e:A\to R$ the structure map, the multiplication map, the inverse map, the unit section map respectively. We say an element $a\in A$ is a group-like element if it's a unit and $m(a)=a\otimes a$. Denote $A^\mathrm{gl}$ the group of group-like elements of $A$.
Using the identification of maps $$A\stackrel{m}{\to}A\otimes_R A\stackrel{(s\circ e,\ \mathrm{id}_A)}{\to}A=A\stackrel{\mathrm{id}_A}{\to}A$$
we know if $a\in A^\mathrm{gl}$, then $s(e(a))\cdot a=a$ hence $s(e(a))=1$ thus $e(a)=1$.
And my problem is, if there is a map of $R$-algebras $\phi :A\to R$ s.t. $\phi (A^\mathrm{gl})=1$. Do we must have $\phi=e$? (It would probably a naive implication of the Cartier duality but I can't see it)
If so, does it imply that for any $R$-algebra $B$ and any two maps of $R$-algebras $\phi_1,\ \phi_2:A\to B$ agreeing on the set of group-like elements $A^\mathrm{gl}$, then $\phi_1 =\phi_2$?
I apologize again for my confusion with definitions. What I mean is. Let $\newcommand{\Zt}{\mathbb Z/3\mathbb Z} \newcommand{\Q}{\mathbb Q}$ $$ A = \{f\colon \mathbb Z/3\mathbb Z \to \Q\} \cong \Q^3 $$ $A$ is a $\Q$-Hopf algebra: the $\Q$-algebra structure is given by pointwise addition and multiplication, and the Hopf part comes from the group structure of $\Zt$, i.e. we identify $A\otimes_\Q A$ with the set of functions $\Zt\times \Zt\to \Q$, via $$ A\otimes A\ni f\otimes g\mapsto \left((a,b)\mapsto f(a)g(b)\right)\in \operatorname{Maps}(\Zt\times \Zt,\Q) $$ Then, $$ m(f)(a,b)=f(a+b)\\ i(f)(a)=f(-a)\\ e(f) = f(0). $$ Suppose $f\colon \Zt\to \Q$ is group-like, i.e. $$ m(f)(a,b) = f(a+b) = f(a)f(b) = (f\otimes f)(a,b) $$ This is equivalent to saying that $f$ is a group homomorphism $\Zt\to \Q^\times$, so the only group-like element is the constant $1$. However, for all three $a\in \Zt$, $f\mapsto f(a)$ is a homomorphism $A\to \Q$ that sends $1$ to $1$.
The group algebra of $\Zt$ is the Cartier dual of $A$ (its standard basis is the dual basis of the standard basis of $A$), and as you say, it is generated by group-like elements.