A formal Dirichlet series over a ring $R$ is associated to a function $a$ from the positive integers to $R$ $$ L(a,s) = \sum_{n=1}^\infty a(n) n^{-s} $$ with addition and multiplication defined by $$ L(a,s) + L(b,s) = \sum_{n=1}^\infty (a+b)(n) n^{-s} $$ $$ L(a,s) \cdot L(b,s) = \sum_{n=1}^\infty (a*b)(n) n^{-s} $$ where $$ (a+b)(n) = a(n)+b(n) $$ is the pointwise sum and $$ (a*b)(n) = \sum_{k|n} a(k)b(n/k) $$ is the Dirichlet convolution of $a$ and $b$.
The formal Dirichlet series form a ring $\Omega$, indeed an $R$-algebra, with the zero function as additive zero element and the function $\delta$ defined by $\delta(1)=1$, $\delta(n)=0$ for $n>1$ (so that $L(\delta,s)=1$) as multiplicative identity. An element of this ring is invertible if $a(1)$ is invertible in $R$. If $R$ is commutative, so is $\Omega$; if $R$ is an integral domain, so is $\Omega$. The non-zero multiplicative functions form a subgroup of the group of units of $\Omega$.
The ring of formal Dirichlet series over $\mathbb{C}$ is isomorphic to a ring of formal power series in countably many variables.
The function $a$ is multiplicative if and only if there is a formal Euler identity beween the Dirichlet series $L(a,s)$ and a formal Euler product over primes $$ L(a,s) = \sum_n a_n n^{-s} = \prod_p (1+a_p p^{-s} + a_{p^2} p^{-2s} + \cdots ) $$ and is totally multiplicative if the Euler product is of the form $$ L(a,s) = \sum_n a_n n^{-s} = \prod_p (1 - a_p p^{-s})^{-1} \ . $$