Here is a idea concerning the classification of finite rings (commutative, unital). Related question: Classification of finite commutative rings.
Every finite ring is a direct product of finite algebras over $\mathbb{Z}/p^n$ for some prime power $p^n$ (Chinese Remainder Theorem). Now fix such a prime power. If $R,S$ are finite algebras over $\mathbb{Z}/p^n$, then $R \otimes_{\mathbb{Z}/p^n} S$ and $R \times S$ are also finite algebras over $\mathbb{Z}/p^n$. If $\mathcal{A}$ is the set of isomorphism classes of finite algebras over $\mathbb{Z}/p^n$, then $\mathcal{A}$ becomes a commutative semiring with addition $\times$, zero element $0$, multiplication $\otimes$ and identity $\mathbb{Z}/p^n$. What is known about the associated ring $\overline{\mathcal{A}}$? Note that the map $\mathcal{A} \to \overline{\mathcal{A}}$ is not injective (I don't think that $\mathcal{A}$ is cancellative), but perhaps this loss of information makes it possible to give a desciription of this ring.
What about varying $n$? For every $k \leq n$ we may regard $\mathcal{A}_k$ as an ideal of $\mathcal{A}_n$, and $R \mapsto R/p^k$ is a projection $\mathcal{A}_n \to \mathcal{A}_k$. Does this help to describe $\mathcal{A}_n$ or $\overline{\mathcal{A}_n}$ recursively?
Best Answer
Not an answer, just a longish comment, I hope correct:
Let's start with the case $p^n=p$, so this is about finite algebras over a field. Let's replace $\mathbb Z/p$ by any field $k$ and ask the same question. The additive monoid of isomorphism classes of such algebras (under cartesian product) does have the cancellation property, because of the way that these rings decompose as products of local rings; it's the free abelian monoid on the set of iso classes of local finite algebras. Let's specialize to the case when $k$ is algebraically closed. Then the tensor product of local algebras is local, so that the semiring is determined by a tensor-product monoid of local algebras, and so is the associated ring.
If you go on to simplify the ring more by inverting these local algebras, you get the integral group ring of some abelian group: the group obtained by adjoining inverses to that monoid of local algebras. I wonder what this group is like. It maps onto the group of positive rational numbers by sending a local algebra to its vector space dimension.
Added: Come to think of it, there's a much bigger invariant than this: To a local algebra A with max ideal m, associate the polynomial $p_A(T)=1 + dim(m/m^2)T+dim(m^2/m^3)T^2+\dots $. This gives a homomorphism onto a much bigger group.