I'm working in the spaces $\mathbb{I}_Z$ and $\mathbb{I}_Q$ of infinite integer/rational sequences that start with $1$:
$$∀x∈\mathbb{I}_Z\ ∀n∈\mathbb{N}:x_0=1,x_n∈\mathbb{Z}$$
$$∀y∈\mathbb{I}_Q\ ∀n∈\mathbb{N}:y_0=1,y_n∈\mathbb{Q}$$
Both sets form abelian groups under discrete convolution defined in the usual way: $(x*y)_n=\sum\limits_{i=0}^nx_iy_{n-i}$ — commutativity and associativity follow directly from the properties of convolution, and one can easily check closure and existence of the neutral and opposite elements. In other words, convolution on these sets behaves a lot like addition: even more to this point, it is also quite easy to see that the equation $\underbrace{x*...*x}_{n\ {\rm times}}=a$ always has a unique solution (it might not be an integer sequence, but it is always a rational one), so it makes sense to write it in the multiplicative notation: $x=a/n$, rather than as a "root". By the way, the usual elementwise addition is basically undefineable in $\mathbb{I}_Z$ or $\mathbb{I}_Q$ since adding the $0$'th terms breaks the closure and trying to fix that usually just leads to more inconsistencies.
This made me wonder whether it is possible to define a "natural" notion of multiplication on $\mathbb{I}_Z$. By "natural" I mean that I would like the operation to be closed in $\mathbb{I}_Z$ (or respectfully in $\mathbb{I}_Q$) and distributive over convolution. It wolud also be nice to have all the good properties like commutativity, associativity, having the identity and inverses; in the best case scenario $\mathbb{I}_Z$ would become a commutative ring with unity or something and $\mathbb{I}_Q$ a field.
My best attempt of reaching this so far is based on the assumption that the first $n$ terms of the product depend only on the first $n$ terms of the operands. This just seemed like a good choice since this is how addition and multiplication usually work in number systems, and it also already works for convolution. Let $x,y,a,b∈\mathbb{I}_Z$ and $$f_n(x,y)=f_n\Big({x_1...x_n\atop y_1...y_n}\Big)=(xy)_n$$ be the $n$'th term of the product $xy$. By the way, the requirement $x_0=y_0=(xy)_0=1$ is pretty obvious, so we'll ignore the $0$'th terms from now on. By the distributive property we can write: $$(x(a*b))_n=(xa*xb)_n=(xa)_n+(xa)_{n-1}(xb)_1+(xa)_{n-2}(xb)_2+...+(xb)_n$$ Plugging $n=1$ and rewriting in terms of $f$ gives: $$f_1\Big({x_1\atop a_1+b_1}\Big)=f_1\Big({x_1\atop a_1}\Big)+f_1\Big({x_1\atop b_1}\Big)$$ The only sensible family of solutions is $f_1\Big({\displaystyle{x_1\atop y_1}}\Big)=k\,x_1y_1$, and it seems like a good idea to choose $k=1$, for example to keep closure and identity. So $f_1\Big({\displaystyle{x_1\atop y_1}}\Big)=x_1y_1$. Continuing with $n=2$: $$f_2\Big({x_1\atop a_1+b_1}\bigg|{x_2\atop a_2+a_1b_1+b_2}\Big)= f_2\Big({x_1\atop a_1}\bigg|{x_2\atop a_2}\Big) +f_1\Big({x_1\atop a_1}\Big)f_1\Big({x_1\atop b_1}\Big) +f_2\Big({x_1\atop b_1}\bigg|{x_2\atop b_2}\Big)$$ turns into $$f_2\Big({x_1\atop a_1+b_1}\bigg|{x_2\atop a_2+a_1b_1+b_2}\Big)= f_2\Big({x_1\atop a_1}\bigg|{x_2\atop a_2}\Big) +x_1^2a_1b_1+f_2\Big({x_1\atop b_1}\bigg|{x_2\atop b_2}\Big)$$ This is a rather tricky functional equation since however you choose $f_2$ to act, it will do so in three different places simultaneously, and you also have to keep it symmetric with respect to the "upper" and "lower" arguments for the sake of commutativity. Luckily, with a little bit of work everything cancels out quite nicely and one can get a family of solutions: $$f_2\Big({x_1\atop y_1}\bigg|{x_2\atop y_2}\Big)=x_1^2y_2+x_2y_1^2-2x_2y_2+k\,x_1y_1$$ Works fine but now I have no idea which value of $k$ to choose. Perhaps $0$ to keep things simpler but how can I be sure? Moreover, for $n=3$ the equation is already a mess: $$f_3\Big({x_1\atop a_1+b_1}\bigg|{x_2\atop a_2+a_1b_1+b_2}\bigg|{x_3\atop a_3+a_2b_1+a_1b_2+b_3}\Big) =f_3\Big({x_1\atop a_1}\bigg|{x_2\atop a_2}\bigg|{x_3\atop a_3}\Big) +f_3\Big({x_1\atop b_1}\bigg|{x_2\atop b_2}\bigg|{x_3\atop b_3}\Big)+$$ $$+x_1a_1(x_1^2b_2+x_2b_1^2-2x_2b_2)+x_1b_1(x_1^2a_2+x_2a_1^2-2x_2a_2)$$ And I'm not even sure whether continuing with this process is a good idea. There's no guarantee that it will even be solvable at each step and that the solution is unique enough. So maybe there's a simpler way of solving the problem? Or a completely different approach? Or some fact I didn't know about or some pattern that I didn't notice? Any help would be greatly appreciated!