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.