The presentation of the formal moduli problems story in Gaitsgory-Rosenblyum A Study in Derived Algebraic Geometry, Vol 2 may be what you are looking for. We review it here (in the case over $\mathrm{Spec}\, k$ for a field $k$ of characteristic zero, that the question concerns):
1. Looping/delooping equivalence in formal DAG
Just like the familiar adjoint equivalence in the homotopy theory of spaces
$$\mathrm B: \mathrm{Grp}_{\mathbb E_1}(\mathcal S)\simeq \mathcal S_*^{\ge 1}:\Omega,$$
there is an analogous adjoint equivalence in formal DAG
$$\mathrm B:\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k) \simeq \mathrm{FMP}_k:\Omega,$$
where the loop space functor is in both cases given by $\Omega X = \mathrm{pt}\times_X \mathrm{pt}$, as per usual in homotopical/$\infty$-categorical things. Note that formal moduli problems are already inherently pointed, by the assumption that $X(k)$ is contractible.
2. Formal groups and Lie algebras
Now, $\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k) = \mathrm{FGrp}_k$ is (an incarnation) of the $\infty$-category of (derived) formal groups over $k$. Thanks to the characteristic zero assumption, there is a further equivalence
$$
\mathrm{Lie}:\mathrm{FGrp}_k \simeq \mathrm{LieAlg}_k : \exp
$$
with derived Lie algebras (as modelled for instance by dg Lie algebras). Just like expected, the derived Lie algebra $\mathfrak g$ corresponding to the formal group $G$ is $\mathfrak g = T_{G, e}$, the tangent fiber at the unit.
3. Formal moduli problems and Lie algebras
The celebrated Lurie-Pridham identification between formal moduli problems and derived Lie algebras is precisely the composite of these two equivalences of $\infty$-categories
$$
\mathrm{FMP}_k\simeq \mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k)=\mathrm{FGrp}_k\simeq \mathrm{LieAlg}_k.
$$
That is
- It sends a formal moduli problem $X$ to
$$
\mathrm{Lie}(\Omega X) = T_{\Omega X, e} = T_{X, x_0}[-1],
$$
where $x_0$ is the base-point of $X$, unique up to a contractible space of choices $X(\kappa)$. This shifted tangent fiber carries a canonical Lie algebra structure coming from (i.e. as the Lie algebra of) the $\mathbb E_1$-group structure of $\Omega X$.
- The inverse equivalence $\Psi: \mathrm{LieAlg}_k \simeq \mathrm{FMP}_k$ is then given by $\Psi(\mathfrak g)= \mathrm B\exp(\mathfrak g)$.
4. Lurie's formula for $\Psi$
You may justifiably complain that this description of the inverse functor $\Psi$ does not look the same as the one in Lurie's writing. Let's see how to get it in that form.
Let's assume that the formal moduli problem $\Psi(\mathfrak g)$ is formally affine (true under some finiteness assumptions on $\mathfrak g$), in the sense that
$$
\mathrm B\exp(\mathfrak g)= (\mathrm{Spf}\,k)/\exp(\mathfrak g)\simeq \mathrm{Spf} \,k^{\mathfrak g}.
$$
Here of course the formal spectrum is the usual functor $\mathrm{Spf}:(\mathrm{CAlg}^\mathrm{aug}_k)^\mathrm{op}\to \mathrm{FMP}_k$. The derived Lie algebra invariants may be computed via the Chevalley-Eilenberg complex, thus $k^\mathfrak g\simeq {C}^*(\mathfrak g)$.
Then for any Artinian $k$-algebra $A$ we have
$$ (\Psi(\mathfrak g))(A) \simeq \mathrm{Map}_{\mathrm{CAlg}^\mathrm{aug}_k}({C}^*(\mathfrak g), A). \qquad \quad(1)
$$
This is why Lurie tells us to consider the "Koszul duality functor" $\mathfrak D: (\mathrm{CAlg}_k^\mathrm{aug})^\mathrm{op}\to \mathrm{LieAlg}_k,$ right-adjoint to the Chevalley-Eilenberg cochains functor. Indeed, (through a little use of the finiteness of $\mathfrak g$) we get
$$(\Psi(\mathfrak g))(A) \simeq \mathrm{Map}_{\mathrm{LieAlg}_k}( \mathfrak D(A), \mathfrak g),\qquad \qquad(2)
$$
which is how Lurie tells us to define $\Psi$.
Note that this is going the other way than the $C^*\dashv\mathfrak D$ adjunction. Indeed, without finiteness assumptions on $\mathfrak g$, the formal moduli problem $\Psi(\mathfrak g)$ will not necessarily be formally affine, and the formula (1) will not necessarily work. On the other hand, as Lurie teaches us, formula (2) will always work.
5. Loose ends
If I understand the original question correctly, this is presenting the formal moduli problem story precisely like the intuitive picture mentioned. Indeed: any formal moduli problem may be written as $X\simeq \mathrm B G$ for a derived formal group $G$, and the formal moduli problem (of mapping into) $\mathrm BG$ is classified by the Lie algebra $\mathfrak g$.
That said, the question makes analogy with the $\mathbb E_n$-algebra version of this story too. I am a little confused about the points raised - in particular, it seems like the following is asserted: a formal moduli problem on $\mathbb E_n$-algebras is represented by an $\mathbb E_n$-algebra. That is, so far as I understand, incorrect. Instead, any FMP on $\mathbb E_n$-algebras in classified by an $\mathbb E_n$-algebra, in a way somewhat analogous to the way that the functor $\mathfrak D$ presents formal moduli problems in the commutative case with Lie algebras.
In particular, there seems to be no need to think about cogroup objects, as the equivalence of $\infty$-categories $\mathrm{FMP}_k^{\mathbb E_n}\simeq \mathrm{Alg}^\mathrm{aug}_{\mathbb E_n}$ is covariant.
$\qquad$
Edit: A question was raised in the comments whether group objects in $\mathrm{FMP}_k^{\mathbb E_n}\simeq \mathrm{Alg}_{\mathbb E_n}^\mathrm{aug}$ correspond to $\mathbb E_{n+1}$-algebras. Unless I am misunderstanding the question, the answer is negative.
A monoid structure (of which a group structure is a particular example of) on an $\mathbb E_n$-algebra $A$ is given by a map $A\times A\to A$, plus coherence data. On the other hand, an additional $\mathbb E_1$-algebra structure on $A$ (which is equivalent to making it into an $\mathbb E_{n+1}$-algebra by Dunn Additivity) is given by a map $A\otimes_k A\to A$, plus coherence data. So a group object in $\mathbb E_n$-algebras, and an $\mathbb E_{n+1}$-algebra are different structures.
This might feel a little weird because we're used to tensor products corresponding to products of schemes. Alas, unlike the contravariant $\mathrm{Spec}$, the equivalence $\Psi:\mathrm{Alg}^{\mathrm{aug}}_{\mathbb E_n}\simeq \mathrm{FMP}_k^{\mathbb E_n}$ is covariant. (PS: another difficulty: the tensor product only becomes the coproduct on the level of $\mathbb E_\infty$-algebras, not $\mathbb E_n$-algebras, so even a contravariant equivalence wouldn't necessarily do the trick).
Rather, the situation is the same as in the $\mathbb E_\infty$-case above: there is an equivalence
$$\mathrm B:\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k^{\mathbb E_n})\simeq \mathrm{FMP}_k^{\mathbb E_n}:\Omega.$$
This is a special case of the following general phenomenon: for any operad $\mathcal O$, the loops functor $\Omega: \mathrm{Alg}_{\mathcal O}(\mathrm{Mod}_k)\to \mathrm{Grp}_{\mathbb E_1}(\mathrm{Alg}_{\mathcal O}(\mathrm{Mod}_k))$ is an equivalence of $\infty$-categories (A Sudy in Derived Algebraic Geometry, Vol 2, Chapter 6, Proposition 1.6.4). The above claims are special cases for $\mathcal O$ the Lie operad and the $\mathbb E_n$-operad respectively.
