This is not an answer but rather extended commentary on Tom
Bachmann's answer. I hope to
address some of Jon's comments on that answer.

# Bar constructions with respect to coproducts compute pushouts

First let me address Tom's point (3), that an instance of the
two-sided bar construction computes pushouts. This is probably
well-known but I had to think about it for a bit, so I'll include an
explanation in case it helps someone else. Also, this explanation
helps later with coherence issues for maps between simplicial objects.

The two-sided bar construction can be defined very generally, for a
monoid $M$ for some monoidal structure $\otimes$ together with
left and right $M$-modules $L$ and $R$. Now consider the special
case when $\otimes$ is given by the coproduct. In this case, the
monoidal structure on $M$ is unique, given by the fold $M \sqcup M
\to M$, and the module structures on $L$ and $R$ are given simply
by morphisms $M \to L$, $M \to R$. And the claim is that the
geometric realization of the bar construction is simply the pushout of
$R \leftarrow M \to L$.

One way to see that is to consider the functor $b : \mathrm{Span} \to
\Delta^\mathrm{op}$ where $\mathrm{Span} = \{ r \leftarrow m \to l
\}$ and $b(m \to l) = (d_0 : [1] \to [0])$ and $b(m \to r) =
(d_1 : [1] \to [0])$. It's not hard to check that the coproduct-based
bar construction $\mathrm{Fun}(\mathrm{Span}, \mathcal{C}) \to
\mathrm{Fun}(\Delta^\mathrm{op}, \mathcal{C})$ is given by left Kan
extension along this functor $b$. The usual argument about left Kan
extending along the composite $\mathrm{Span} \to \Delta^\mathrm{op}
\to \ast$, shows that a span and its corresponding bar construction
have the same colimit.

# Defining the relevant map of simplicial $A$-modules

Instead of trying to check that Tom's maps are compatible with faces
and degeneracies, let's try to build simplicial maps wholesale. I'll
use $F : \mathrm{Mod}_A \to \mathrm{Alg}^{E_n}_{A}$ for the free
$E_n$-$A$-algebra functor defined on $A$-module spectra, and
$U : \mathrm{Alg}^{E_n}_{A} \to \mathrm{Mod}_A$ for the
corresponding forgetful functor.

The key ingredient in Tom's argument is a simplicial $A$-module
$B_{\bullet}$ whose geometric realization is $U(A/\!/\alpha)$, the
underlying $A$-module of the $E_n$-$A$-algebra in Jon's
question. By the previous section, the simplicial
$E_n$-$A$-algebra $\mathcal{B}_{\bullet} :=
\mathrm{Lan}_b(A\overset{\overline{0}}\leftarrow
{E_n}(\Sigma^kA)\overset{\overline{\alpha}}\to A)$ has geometric
realization given by the pushout of that span, $A /\!/ \alpha$.
Since the forgetful functor $U$ preserves geometric realizations,
the simplicial $A$-module $B_{\bullet} := U \circ
\mathcal{B}_{\bullet}$ will have $U(A /\!/ \alpha)$ as geometric
realization.

Now let $C_{\bullet} := \mathrm{Lan}_b(0 \leftarrow \Sigma^k A \to
A)$ be the coproduct-based bar construction in $\mathrm{Mod}_A$. We
want to define a morphism of simplicial $A$-modules $C_{\bullet}
\to B_{\bullet} = U \circ \mathcal{B}_{\bullet}$, or equivalently a
morphism of simplicial $E_n$-$A$-algebras $F \circ C_{\bullet}
\to \mathcal{B}_{\bullet}$. Now, since $F$ is a left adjoint it
preserves the left Kan extension defining $C_{\bullet}$, so $F
\circ C_{\bullet}$ is the bar construction for the span $F(0)
\leftarrow F(\Sigma^k A) \to F(A)$. The following diagram is a
natural transformation between that span and the one defining
$\mathcal{B}_{\bullet}\require{AMScd}$:
$$\begin{CD}
F(0) @<{F(0)}<< F(\Sigma^k A) @>{F(\alpha)}>> F(A) \\
@V{\mathrm{id}}VV @V{\mathrm{id}}VV @V{\mu_A}VV \\
A @<{\bar{0}}<< F(\Sigma^k A) @>{\bar{\alpha}}>> A \\
\end{CD}$$
This map of spans induces a simplicial map between their respective
bar constructions.

# The rest of the argument

I think that now the rest of Tom's argument runs fine: let
$B'_{\bullet}$ be the cofibre in simplicial $A$-modules of
$C_{\bullet} \to B_{\bullet}$. Now, without worrying about
compatibility with faces and degeneracies, you can identify for each
$n$ the map $C_n \to B_n$ to describe $B'_n$ and see that
$B'_n$ is $2k$-connective.

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