Let $\varphi : \Bbb{N} \to \Bbb{N}$ be a completely additive arithmetic function, i.e. $\varphi(nm) = \varphi(n) + \varphi(m)$ for all $n, m \in \Bbb{N}$.
For example, $\Omega(n) = $ the total number of prime factors of $n$ including multiplicities.
Let $U \subset \Bbb{Z}^{\Bbb{N}}$ be a subset of integer sequences defined by $U = \{ u \in \Bbb{Z}^{\Bbb{N}} : \sum_{n \geq 1} u_n n = 0, \text{supp}(u) \lt \infty\}$. That is to say vectors in $U$ represent $\Bbb{Z}$-linear relationships among some finite subset of natural numbers. $U$ is clearly an abelian group, as adding $0$ to $0$ equals $0$ or another relation among the integers. The set of linear relationships is in bijection with $U$.
Let $(u + v)_n = u_n + v_n$ be defined componentwise. However, instead of componentwise multiplication, we define a convolution-type product - this is why $U$ may be isomorphic to something else in arithmetic function theory.
To make an elegant definition of multiplication, we first define the abelian groups $A_i = \{ u \in U : u_n = 0, \ \forall n \in \Bbb{N}$ such that $\varphi(n) \neq i \}$.
Then such subsets of $U$ are indeed abelian groups since the condition $\varphi(n) \neq i \implies u_n = 0$ works strictly with the indices. So the values of the sequence are free to be any value $u_n \in \Bbb{Z}$.
Also notice that these abelian groups have pairwise intersection $A_i \cap A_j = 0 \in U$ for all $i \neq j$. This is because of the fact that two distinct inverse images of a function are disjoint, here $\varphi^{-1}(i) \cap \varphi^{-1}(j) = \varnothing$ and so form a partition of $\Bbb{N}$. In the case of $\varphi = \Omega$, you can write this as $\Bbb{N} = \{1\} \uplus \Bbb{P} \uplus \Bbb{P}^{2} \uplus \dots$ where $\Bbb{P} \subset \Bbb{N}$ is the set of prime numbers and $\Bbb{P}^k = \{ q_1 \cdots q_k : q_j \in \Bbb{P}\}$ is the set of products of $k$ primes.
Now, since we're defining a graded ring structure any two general elements $x, y \in U$ that are nonzero belong to a unique $A_i, A_j$, respectively.
Thus let $x, y \in U$ and $x \in A_i, y \in A_j$. Then define the convolution product for the ring as:
$$ (xy)_n = \sum_{ab = n \\ \varphi(a) = i \\ \varphi(b) = j} x_a y_b $$
This is essentially derived from multiplying two linear relations and taking their coefficients:
$$ 0 = (\sum_{n \geq 1} x_n n)(\sum_{n \geq 1} y_n n) $$
For example: multiplying $\sum_{p \text{ prime}} x_p p = 0$ and $\sum_{p, q \text{ prime}} y_p pq = 0$, corresponding to elements $x \in A_1, y \in A_2$, will clearly yield a relation corresponding to an element of $A_3$, however we have to keep careful track of where the coefficients end up and that is the purpose of the above formula.
You can see by looking at the formulas for $+$ and $\cdot$ that the distributive law should hold. Associativity of $\cdot$ is brought over from associativity in $\Bbb{Z}$, when we multiply two linear relations and take their coefficients.
Thus, we have that for all $i,j \in \Bbb{N} \cup 0$, $A_i A_j \subset A_{i + j}$, each $A_i$ is an abelian group and $A_i \cap A_j = 0$. That is precisely the definition of a graded ring:
$$ R_{\varphi} = \bigoplus_{n \geq 0 } A_n $$
Note that $A_0 := \Bbb{Z}$ must be defined. See arguments in comments of an answer below. This ensures unitality of the ring.
Thus, let $R_{\varphi}$ be called the graded ring associated with the completely additive arithmetic function $\varphi$.
Note that for two completely additive arithmetic functions $\varphi, \psi$, $R_{\varphi}$ need not be isomorphic to $R_{\psi}$ as rings, though their underlying additive abelian groups $U$ are equal.
My question is, can you see whether $R_{\varphi}$ is isomorphic to some ring already known about?
My next question is: how would you generalize this, such as replacing the coefficients with $\Bbb{Q}$ or some integral domain, replacing $\Bbb{N}$ with $\Bbb{Z}$ or a group, so that $\varphi$ becomes an additive group character, etc. ?
Note that any such $\varphi : \Bbb{N} \to \Bbb{N} \cup 0$ extends nicely to all of $\Bbb{Z}$ by setting $\varphi(-n) = \varphi(|n|)$ and then it extends nicely to all of $\Bbb{Q}^{\times}$ via $\varphi(x/y) := \varphi(x) - \varphi(y)$ where it becomes an abelian group homomorphism onto $(\Bbb{Z}, +)$.
Without the $U$ condition your ring is that of Dirichlet polynomials with integer coefficients (more generally formal Dirichlet series), decomposed (graded by the values taken by $\varphi$) as $$\sum_{n\ge 1} a_n n^{-s}= \sum_j \sum_{n\ge 1,\varphi(n)=j} a_n n^{-s}$$
This ring is $\Bbb{Z}[2^{-s},3^{-s},5^{-s},\ldots]\cong \Bbb{Z}[X_1,X_2,X_3,\ldots]$ (for the formal Dirichlet series take the formal series in the $X_j$).
Don't forget that the simplest additive function (and the natural grading on Dirichlet series) is $\log n$.
With the $U$ condition then $1$ is not in your ring and everything is not quite natural (the Dirichlet polynomials whose every graded piece vanish at $-1$..).
Grading by $\Omega(n)$ is somewhat natural in $\zeta(s)=\exp(\sum_{p^k} p^{-sk}/k)$.