One type of integer divisor homology - what does this one measure? The set of functions $\{ (d \mid \cdot) : d \in \Bbb{N}\}$ where $$(d\mid x) = \begin{cases} 1, \text{ if } d \mid x, \\ 0, \text{ if } d \nmid x \end{cases}, \ \forall x \in \Bbb{Z}$$
are the "divisor indicator functions" and they generate a subspace of the $\Bbb{Z}$-module $M = \{ f : \Bbb{N} \to \Bbb{Z}, f \text{ any function}\}$.  What subspace, I don't know, so we'll just say the one that they generate.  The $(d\mid x)$ are easily seen to be linearly independent (see comment).
Now, consider $\Bbb{Z}$-linear combinations of the $(d\mid \cdot)$'s.
Define $M_n = \{ f \in M : f(x) = \sum_{d \in \Bbb{N}} c_d (d \mid x), \ c_d \in \Bbb{Z} \}, n \geq 0$ where we do not require all but finitely many $c_d$ to be nonzero as long as the domain of the functions we consider does not include $0$, since $(d \mid 0) = 1$ always so there would be an infinite sum of coefficients which may not converge.

Now, define $\partial = \partial_n : M_n\to M_{n-1}$ for all $n \geq 1$ in the obvious way:
Let $d = q_1 \cdots q_n$ be such that $q_1 \leq q_2 \leq \dots \leq q_n$ is the standard ordering of the primes comprising $d$.
$$
\partial (d \mid \cdot) = \partial(q_1 \cdots q_n \mid \cdot) \\ 
= \sum_{i = 1}^n (-1)^i (q_1 \cdots \widehat{q_i} \cdots q_n \mid \cdot)
$$
and extend each $\partial$ in the unique $\Bbb{Z}$-linear way.
Then for example:
$$
\partial(q_1q_2q_3q_4 \mid \cdot) = -(q_2q_3q_4\mid \cdot) + (q_1q_3q_4\mid \cdot) -(q_1q_2q_4\mid \cdot) + (q_1q_2q_3\mid \cdot)
$$
so that:
$$\partial\circ \partial(q_1q_2q_3q_4\mid \cdot) \\
= (q_3q_4 \mid \cdot) - (q_2q_4\mid \cdot) + (q_2q_3 \mid \cdot) \\
- (q_3q_4\mid \cdot) + (q_1q_4\mid \cdot) -(q_1q_3 \mid \cdot) 
+(q_2q_4 \mid \cdot) - \dots \\
= 0
$$
Since we do have a chain complex: $\cdots \xrightarrow{\partial} M_2 \xrightarrow{\partial} M_1 \xrightarrow{\partial} M_0$
Question.

What does the $n$th homology module measure?


Attempt.
Let $R$ be a ring with $1$ and define $M_n = \{ f : \Bbb{N} \to R \vert f(x) = \sum_{d \in \Bbb{N}, \\ \Omega(d) = n} c_d(d \mid x), c_d \in R \}$.
Then $\partial_n$ defined in the question post is still relevant.  Let $p, q, p_i, q_i$ always mean a prime number, except $p_i$ is the standard $i$th prime number and $q_i$ is the $i$th prime number in some contextual list of prime numbers.  Fix a linear ordering on all prime numbers wherever they occur.  When we write $q_1 \cdots q_n$ we will implicitly mean that $q_1 \leq \dots \leq q_n$ under this chosen ordering.
Define $Z_n = \ker \partial_n$ where $\partial_n : M_n \to M_{n-1}$ for all $n \geq 1$.  Similarly define $B_n = \partial_n(M_n)$.
$$
Z_n = \{ f \in M_n : \partial(f)(x)=\sum_{d \in \Bbb{N} \\ \Omega(d) = n} f_d \sum_{q_i \mid d \\ i = 1..\Omega(d)}(-1)^i(\frac{d}{q_i} \mid x) = 0, \ \forall x \in \Bbb{N}\}
$$
where empty summations (no $q \mid d$ exists) are taken to be $0$.

$$
B_{n} = \{ \partial(f) = \sum_{d \in \Bbb{N} \\ \Omega(d) = n+1} f_d \sum_{q_i \mid d \\ i = 1..\Omega(d)}(-1)^i(\frac{d}{q_i} \mid x) : f \in M_{n+1}\}
$$

A boundary map is $\partial(f)(x) = \sum_{d \in \Bbb{N}} f_d\sum_{q_i \mid d}(-1)^i(\frac{d}{q_i} \mid x)$ so:
$$
B_1 = \partial(M_2) = \sum_{p,q\in \Bbb{P} \\ p\leq q} \Bbb{Z}((q\mid \cdot) - (p\mid \cdot)) \\
B_2 = \partial(M_3) = \sum_{p,q,r\in \Bbb{P} \\ p\leq q \leq r} \Bbb{Z}((qr\mid \cdot) -(pr\mid \cdot) +(pq\mid\cdot)) \\
Z_1 = \ker \partial_1 = \{ f \in M_1 : \sum_{p \in \Bbb{P}} f_p = 0\} \\
Z_2= \ker \partial_2 = \{ f \in M_2 : \sum_{p\leq q} f_{pq}((q\mid \cdot) - (p\mid \cdot)) = 0\}
$$
 A: I wanted to mention a few things I've noticed. I'm assuming $0 \notin \mathbb N$. The function $(0|x)$ is a bit more useless than I thought; since $0 | x$ if and only if $x = 0$, $(0|x) = \delta_{0x}$ is a Kronecker delta at $0$, so it's not very expressive. Removing it makes the theory more elegant.
Let $K$ be a ring. The set $\mathrm{Hom}_{\mathbf{Set}}(\mathbb Z, K) = \{ f : \mathbb Z \to K \}$ is a ring under pointwise addition and multiplication and the constant function equal to $1$ serves as an identity. The $\mathbb Z$-submodule $\mathrm{DFR}(\mathbb Z,K) = \langle \{ (d | x) \, | \, d \in \mathbb N\} \rangle_K \subseteq \mathrm{Hom}_{\mathbf{Set}}(\mathbb Z, K)$ is a unital subring (I'm calling it the divisor function ring or $\mathrm{DFR}$ for short); the identity of the ring $\mathrm{Hom}_{\mathrm{Set}}(\mathbb Z, K)$ is equal to $(1|x)$, and if $d,d' \in \mathbb N$, we have $(d|x)(d'|x) = ([d,d']|x)$ where $[d,d']$ denotes the least common multiple of $d$ and $d'$ (because $d | x$ and $d' | x$ if and only if $[d,d'] | x$).
The ring $\mathrm{DFR}(\mathbb Z,K)$ is a proper subring of $\mathrm{Hom}_{\mathrm{Set}}(\mathbb Z, K)$ (a very small one even). For suppose $f \in \mathrm{DFR}(\mathbb Z,K)$, so that $f(x) = \sum_{d=1}^n c_d (d | x)$. Let $D = \underset{c_d \neq 0}{\prod_{d=1}^n d}$. For $x \in \mathbb N$ with $\mathrm{gcd}(D,x) = 1$, we have $f(x) = c_1$. Elements of the divisor function ring are "supported" on finitely many divisors (the "support" here would be the set of $d \in \mathbb N$ such that $f(d) \neq c_1$).
By assumption, for all $f \in \mathrm{DFR}(\mathbb Z,K)$, we have $f(0) = \sum_{n \in \mathbb N} c_n$ (because $d | 0$ For all $d \in \mathbb N$) and $f(-x) = f(x)$, so really $f$ only contains the data of a function $f : \mathbb N \to \mathbb Z$. For other rings than $\mathbb Z$ where this construction might work, this "simplification" of restricting $f$ from $\mathbb Z$ to $\mathbb N$ might not make much sense because let's say $K$ is a number ring, the quotient set $K / \{ \text{equivalence up to a unit} \}$ does not have a meaningful structure like $\mathbb N$ does. So I think $\mathbb Z$ is the right domain to use for this problem, not $\mathbb N$. In general, you could work with say a number ring or a UFD.
Using the Möbius inversion formula, we can re-write $f(n) = \sum_{d=1}^n c_d (d|n) = \sum_{d | n} c_d$ as follows:
$$
c_n = \sum_{d | n} \mu(d) f(n/d).
$$
In particular, if $f(n) = 0$ for all $n \in \mathbb N$, then $c_d = 0$ for all $d \in \mathbb N$, so the functions $(d|x)$ are linearly independent, as you claimed.

I'm interested so I'll come back to this answer and edit it, but I need to give more thought to the homology part. I think in particular you might want to define
$$
M_n = \left\{ f = \underset{\Omega(d) = n}{\sum_{d \in \mathbb N}} c_d (d|x) \in \mathrm{Hom}_{\mathrm{Set}}(\mathbb Z, K) \right\}
$$
with the differential the way it is defined now, i.e. that for $\delta_n : M_n \to M_{n-1}$ and $d = p_1 \cdots p_n$ where $p_1 \le \cdots \le p_n$ are all the primes (counted with multiplicity) in the factorization of $d$, and $\delta((d|x)) = \sum_{i=1}^n (-1)^i (d/p_i|x)$.
The thing that makes me curious is that ordering on the primes, it doesn't generalize well to other rings like number rings or UFDs. It's not that I want to generalize, it's that things that generalize well usually are more relevant in an algebraic setting.
EDIT: So, I think I can show that the first two homology groups are zero, i.e. this complex is exact at $M_0$ and $M_1$ (with my definition just above).
We have $M_0 = \langle (1|x) \rangle_K$, and $\delta_0((1|x)) = 0$, so $\delta_0 = 0$ (this is what we expect). We also have $M_1 = \langle (p|x) \, | \, p \text{ prime } \rangle_K$, and $\delta_1((p|x)) = -(1|x)$, so $M_1 \to M_0$ is surjective, and we therefore get exactness at $M_0$.
Now we also have $M_2 = \langle (pq|x) \, | \, p,q \text{ prime, } p \le q \rangle_K$. The map $\delta_2 : M_2 \to M_1$ satisfies $\delta_2((pq|x)) = (q|x) - (p|x)$, so in the homology group $H^1(M_{\bullet})$, we have $(p|x) \equiv (q|x)$. Suppose that $f \in M_1$ satisfies $\delta_1(f) = 0$, i.e.
$$
f(x) = \sum_{i=1}^n c_i(p_i|x), \qquad \delta_1(f)(x) = -\left(\sum_{i=1}^n c_i \right)(1|x) = 0.
$$
The latter condition is equivalent to $\sum_{i=1}^n c_i = 0$ (just plug $x=1$ in the equation $\delta_1(f)(x) = 0$). But in the homology group, $(p_i|x) \equiv (p_j|x)$ because $(p_j|x) - (p_i|x) = \delta_2((p_ip_j|x))$ when $p_i \le p_j$, so we have
$$
f(x) \equiv \left( \sum_{i=1}^n c_i \right)(p_1|x) \equiv 0.
$$
Hope that helps,
A: I have a proof that this complex we defined is exact using simplicial homology. The set $\mathbb N$ forms an abstract simplicial complex in the following fashion. Faces are integers, and the dimension of the face $d \in \mathbb N$ is given by $\Omega(d)$. A face $d$ is included in a face $e$ if and only if $d$ divides $e$ (it's my chosen definition). The face corresponding to $d = q_1 \cdots q_n$ is actually a quotient of the $(n-1)$-dimensional simplex; the quotient map is given by considering a set $\{v_1,\cdots,v_n\}$ of vertices and for $i_1 < \cdots < i_k$, sending the face $(v_{i_1},\cdots,v_{i_k})$ of $\Delta_{n-1}$ to $q_{i_1} \cdots q_{i_k}$, which divides $d$. (The map may not be bijective when some primes $q_i$ are equal.)
By identifying $d$ with $(d|x)$, we see that the set $\{(d|x) \, | \, d \in \mathbb N \}$ also forms an abstract simplicial complex isomorphic to that of $\mathbb N$, the face $d$ in $\mathbb N$ corresponding to the face $(d|x)$. We denote the face corresponding to $(d|x)$ by $\Delta(d)$, and the whole simplicial complex by $\Delta(\mathbb N)$.
Consider an integer $d = q_1 \cdots q_n$ with $\Omega(d) = n$ and $q_1 \le \cdots \le q_n$ all prime numbers. We claim that the function $(d|x)$, as an element of $H_n(M_{\bullet})$, is zero. For this, consider the abstract simplicial complex associated to $(d|x)$. We thus get a map of abstract simplicial complexes $\varphi : \Delta_{n-1} \to \Delta(d)$ given by setting
$$
\forall k,  1 \le k \le n, \quad 1 \le i_1 < \cdots < i_k \le n, \quad (v_{i_1}, \cdots, v_{i_k}) \mapsto (q_{i_1} \cdots q_{i_k}|x).
$$
as described earlier. Since homology is functorial, this gives an induced map of abelian groups $H_n(\varphi) : H_n(\Delta_{n-1}) \to H_n(\Delta(\mathbb N))$. We see also that $(d|x) = \varphi((v_1,\cdots,v_n))$, so $(d|x) = H_n(\varphi)((v_1,\cdots,v_n)) \pmod{B_n(\Delta(\mathbb N))\,}$. But $\Delta_{n-1}$ is contractible as a topological space, so its singular homology (which agrees with its simplicial homology) is zero. Therefore, $(d|x)$ is in the image of the zero map when considered as an element of the homology group, and is therefore zero also.
P.S. : I was computing with $K = \mathbb Z$, but $H_n(M_{\bullet},K) \simeq H_n(M_{\bullet}) \otimes_{\mathbb Z}K = 0$ too, so it doesn't change much.
Hope that helps,
