For an instance, I want to know how to call a group $G$ such that $$ \mathrm{Gp} \vdash A \iff G \models A $$ for any sentence $A$ of the first-order logic, where $\mathrm{Gp}$ is the group axioms.
That is, I am wondering what the name of a model of a given axioms is, which satisfies only the theory of the axioms in a specific formal system.
No such group exists.
For every structure $M$ and every sentence $\varphi$ in the appropriate language, either $M\models\varphi$ or $M\models\neg\varphi$. This means that, since the theory of groups is not complete, every group $G$ will satisfy some sentence which isn't in the theory of groups. There are lots of examples:
Either $G\models\forall x,y(x=y)$ or $G\models\exists x,y(x\not=y)$, even though neither of those sentences is a consequence of the group axioms.
Either $G\models\forall x,y(x*y=y*x)$ or $G\models\exists x,y(x*y\not=y*x)$, even though neither of those sentences is a consequence of the group axioms.
And so on. Indeed, the theory of a structure is always complete - so we only have $$\forall\varphi(M\models\varphi\iff T\vdash\varphi)$$ in case $M\models T$ and $T$ is complete.
Completeness can go away once we consider classes of structures as opposed to individual structures. For a class of structures $\mathbb{K}$, let $$Th(\mathbb{K})=\{\varphi: \forall M\in\mathbb{K}(M\models\varphi)\}=\bigcap_{M\in\mathbb{K}}Th(M).$$ In general $Th(\mathbb{K})$ is not complete, and $Th(\mathbb{Groups})$ is as expected exactly the set of sentences which are consequences of the group axioms.
But this is a very different situation.