Is this almost-cosimplicial object familiar? I have a sequence $X_0,X_1,\ldots$ of Abelian groups, along with face maps $d_0,\ldots, d_n: X_{n-1}\to X_n$ which satisfy the standard cosimplicial identities EXCEPT at the very bottom, where in $\mathrm{Hom}(X_0, X_2)$, I have
$$d_2d_0 \ne d_0d_1$$
However, I have one extra map $d_\infty:X_1\to X_2$ which satisfies the following relations in $\mathrm{Hom}(X_0,X_2)$
$$
d_\infty d_0 = d_2d_0
$$
$$
d_\infty d_1 = d_0d_1
$$
and the following relations in $\mathrm{Hom}(X_1,X_3)$
$$
d_0d_\infty = d_1d_\infty
$$
$$
d_2d_\infty = d_3d_\infty
$$
By hand, I can build a chain complex out of $X_\cdot$, adding a $d_\infty$ to the usual cosimplicial differential. The relations above imply that $d^2=0$.

Is this familiar to anyone?

At first I had hoped that I just had the "wrong" face maps and if I changed around my face maps $X_1\to X_2$ in a clever enough way that this would become a regular (semi)-cosimplicial Abelian group. But I don't see it. 
Another thought I had is that I have many other $d_\infty$ maps that I just haven't seen yet because I don't know the relations that they should satisfy and so am not looking in the right place.
I also noticed that $d_\infty$ satisfies the relations I'd expect from the map $[1]=\lbrace 0,1\rbrace$ to $[2]=\lbrace 0,1,2\rbrace$ which is the constant map to $1$. But this seems not to lead anywhere because of the failure of the face identities at the bottom; i.e., the maps $X_0\to X_1$ and $X_1\to X_2$ don't act like regular injections of ordered sets.
 A: I will also talk about simplicial things rather than cosimplicial things.  Furthermore, I will say "semi-trapezoidal" instead of "almost-simplicial". "Semi-" because there is no codegeneracy maps, and "trapezoidal" because an element $x$ of $X_2$ can be visualized as a trapezoid with four edges $d_2(x), d_\infty(x), d_0(x)$ and $d_1(x)$.
There is a way to make a semi-trapezoidal set into a semi-simplicial set. Given a semi-trapezoidal set $X$ define a semi-simplicial set $A$ setting
$A_0 = X_0, A_1 = X_1 + 2X_2, A_2 = 4X_2, A_3 = X_3 + X_2$ and $A_n = X_n$ for $n > 3.$
Roughly speaking, what we are doing is: We leave the $0$-simplices unchanged. For $1$-simplices we take those of $X$, and add to them the diagonals of the trapezoids (directed in a certain way). For the set of 2-simplices we take the set of triangles in which trapezoids are divided by their diagonals (there are four triangle for each trapezoid). Each trapezoid becomes a 3-simplex of $A$, as well as each original 3-simplex of $X$ becomes a 3-simplex of $A$. More concretely:  
Denote the two embeddings of $X_2 \rightarrow A_1$ by $i_a$ and $i_b$, and their respective images by $A_{a}$ and $A_{b}$. Denote the embeddings $X_2 \rightarrow A_2$ by $i_{aa}, i_{ab}, i_{ba}, i_{bb}$ and their respective images by $A_{aa}, A_{ab}, A_{ba}, A_{bb}$. Define the face maps $e$ of $A$ by:
On $X_1 \subset A_1$: $e_0 = d_0, e_1 = d_1$. 
On $A_{a}$: $e_0 = d_0d_0 = d_0d_1$, $e_1 = d_0d_\infty = d_0d_2$
On $A_{b}$: $e_0 = d_1d_\infty = d_1d_0$, $e_1 = d_1d_2 = d_1d_1$
On $A_{aa}$: $e_0 = i_a, e_1 = d_2, e_2 = d_2$, and similarly on $A_{ab}$, $A_{ba}$, $A_{bb}$.
On $X_2 \subset A_3$: $e_0 = i_{ab}, e_1 = i_{bb}, e_2 = i_{aa}, e_3 = i_{ba}$.
On $X_3 \subset A_3$: $e_0 = i_{ab}d_0, e_1 = i_{bb}d_1, e_2 = i_{aa}d_2, e_3 = i_{ba}d_3$.
On $A_n$: $e_i = d_i$.
This construction extends to a functor and works for simplicial objects in a category with coproduct.
If we are in an abelian category, we can define chain complexes $CX$ and $CA$. There is a morphism $CX \rightarrow CA$ whose component $CX_2 \rightarrow CA_2$ is defined by $i_{aa} + i_{ab}$, which should be a quasi-isomorphism if the intuition is not wrong. 
A: If I were putting money on it, I would wager that the answer to the question you highlighted is "No."  But here's an extended remark that may or may not be helpful.
It's hard for me to think about "co" things, so let me talk about an object which is almost-simplicial in the way that yours is almost-cosimplicial.  I mean I have sets $C_0,C_1,\dots$ and boundary maps $C_i \to C_{i-1}$ that satisfy the usual simplicial relations except in low degrees.
Here's one way that almost-simplicial sets can arise from some categories.  Recall that the usual way to get a simplicial set from a category $C$ is to set $C_0 = \mathrm{Op}(C)$ and $C_1 = \mathrm{Mor}(C)$ and $C_i=\lbrace i$-tuples of composable morphisms in $C\rbrace$. This is not what I'm going to do.  Rather, I want to keep those $C_0$ and $C_1$, but I'd like to think of $C_2$ as some set of zig-zags $x \rightarrow y_1 \leftarrow y_2 \rightarrow z$, and similarly I want $C_i$ to have as its elements some zig-zags with $i$ forward arrows and $i-1$ backwards arrows.
I don't want to consider all zig-zags, but just some of them.  I don't know the general conditions, but the idea is that sometimes, in some categories, you have a zig-zag which nevertheless has a well-defined composition $x \to z$.  This can happen when the zig-zag came with some extra conditions.  An imperfect analogy is to think of $x \to y_1 \leftarrow y_2 \to z$ as a "thickening" of $x \to z$ in the sense of Stolz and Teichner, Traces in monoidal categories.  I guess another case is if the forward arrows are morphisms in a model category that are simultaneously fibrations and cofibrations, and the backward arrows are weak equivalences, then various factorization and lifting properties allow you to do the composition?
Anyway, if you can set up such a situation, then the face maps for most of the $C_i$s just work out for the same reason they always do --- because of associativity of composition.  But in low $i$, you are trying to send a zig-zag $x \to y_1 \leftarrow y_2 \to z$ to either object $y_1$ or $y_2$.
Finally, in this setting, your "extra" face map $d_\infty$ is the one that takes in a zig-zag and returns the arrow $y_2 \to y_1$.
