What is the full algebra of Euler products?
Let the algebra of Euler products be defined by the set of functions which are multiplicative on the domain of integers:
$A=\{a(n):a(n_1\cdot n_2)= a(n_1)\cdot a(n_2)\forall n\in\mathbb{Z}\}$
Complemented by a description of its possible operations - to be determined. And further restricted by the modular relation which states that any two $a\in A$ which give the same solution to:
$\displaystyle f_p(s)=\sum_{n=1}^{\infty}a_p(s)n^{-s}$
are considered equivalent:
$a_{p1}\sim a_{p2}\iff f_{p1}=f_{p2}$
Beginning without reference to the Euler product $f(s)$, only considering the algebra of $A$: multiplicative functions of integers, this would appear to at least be a ring, since we can easily define $a(n)=n^z\mid z\in\mathbb{Z}$ as the integers of that ring with $n^{z_1+z_2}$ the additive function and $n^{z_1\times z_2}$ multiplication.
By way of example we can say that our zero is $a_0(n)=n^0$ and our additive unit is $a_1(n)=n^1$ and by addition we have $a_{2}(n)=a_{1+1}(n)=n^{1+1}=n^2$ so we can therefore define addition of $\sum_{n=1}^\infty n\cdot n^{-s}$ to itself as yielding $\sum_{n=1}^\infty n^2\cdot n^{-s}$
In fact unless I'm mistaken the algebra of $A$ alone is at least a field since the function $n^z$ still yields a unique function $a(n)=n^z$ for every real number $z$.
It would seem the algebra of $A$ is richer still since we can introduce further forms beyond $n^z$ such as $a(n)=(-1)^n$, modular forms such as occurs in Leibniz's formula for $\pi$ and even valuations such as $2^{v_2(x)}$ and $\lvert \cdot\rvert_p$ which are also multiplicative.
It would seem that even the lesser task of defining the full algebra of multiplicative functions would be a powerful tool in analysing the Zeta function.
If one were to define the full algebra of Euler products - a (possibly) modular algebra restricted to end-results which give a unique Euler expansions of the Dirichlet function - certainly an endeavour well beyond my very limited capabilities - surely access to the full gamut of Euler expansions of the Zeta function, and the algebra which governs them, would give us significant greater power than we currently have to analyse its full properties?
In short, there is the only way to obtain a ring of multiplicative Dirichlet series (up to minor modifications) $$\log F(s) = \log \prod_p (1+\sum_{k \ge 1} a_F(p^k) p^{-sk}) = \sum_{p^k} \frac{b_F(p^k)}{k} p^{-sk}$$
$$\log F(s)+\log G(s) = \sum_{p^k} \frac{b_F(p^k)+b_G(p^k)}{k} p^{-sk}$$ $$\log F(s) \otimes \log G(s) = \sum_{p^k} \frac{b_F(p^k)b_G(p^k)}{k} p^{-sk}$$
If $F,G$ both comes from an automorphic form then it is natural to replace $\otimes$ by the Rankin-Selberg convolution $$F(s) \times G(s)= \prod_p (1+\sum_{k \ge 1}a_F(p^k)a_G(p^k) p^{-sk})$$