## Preliminaries

First part of your question doesn't use the bialgebra structure. That is, you have a space of functions on countable many points which I'll denote $A = \mathbb C_1\times \mathbb C_2\times\cdots \times\mathbb C_n\times\cdots$ equipped with $+$ and $\times$ pointwise. You would like to classify all algebra maps $\varphi: A\to \mathbb C$.

To start, note that idempotents must go to idempotents, of which $\mathbb C$ has only two: $0$ and $1$. Moreover, $1\times1\times\cdots \mapsto 1$ (unless the whole map is trivial).

## Classification of points

Consider characteristic functions $\lambda_n$ which take value from point labeled $n$. As $\lambda_n^2 = \lambda_n$, we must have $\varphi(\lambda_n) = 0$ or $1$. There are three cases:

- there are two indices $i, j$ such that $\varphi(\lambda_i) = \varphi(\lambda_j) = 1$. This is impossible: $0 = \varphi(\lambda_i\lambda_j) = 1$.
- there exists exactly one index $i$ for which $\varphi(\lambda_i) = 1$. This implies $\varphi(f) = \varphi((1-\lambda_i)f + \lambda_if) = f(i)$.
- all functions $\lambda_i$, and, therefore, all finite sums from $A$, lie in the kernel of $\varphi$.

The maps of the latter type are indeed the "extra" points. Note, however, that they are "wild" and can be easily killed by some extra finitness assumptions, for example the following "limit of zeroes" property: if $\varphi$ restricted to all finite subsets is 0, then $\varphi$ is 0.

## Wild maps

You can construct **examples of these wild maps** using the axiom of choice. To do that, make the elements of the form $a\times a \times \cdots$ and all their finite modifications to map to $a$; denote those elements $A_0$. Next, select an arbitrary $T_1\in A$ which is transcendental over $A_0$ and map it to arbitrary $t_1\in \mathbb C$. This will fix all elements that lie in algebraic closure $A_1 = A_0(T_1)$. Do that again for $T_2$ and repeat until you have nothing left. At each step you're producing a correct map $A_n \to \mathbb C$, so you get the final map $A\to \mathbb C$ as the limit of those.

Conversely, any wild map can be constructed by the above process, assuming all necessary set-theoretic things. So, the answer is, the wild points are classified by maps from this terribly non-constructive sequence of $T_i$ (of cardinality the same as $A$, that is, continuum) to $\mathbb C$. Those are again in cardinality of **continuum**.

## Monoid structure

One now is reminded that $A$ came with a natural basis enumerated by numbers $(n_1, n_2, \dots, n_k, \dots) \in \mathbb Z^+\oplus\mathbb Z^+\oplus\cdots\oplus\mathbb Z^+\oplus\cdots$ (which is isomorphic to $\mathbb Z_{>0}^\times$). Therefore, it should be possible to add any two points (denoted $\oplus$). The following properties are clear:

- regular points add normally as $n_k = n_k' + n_k''$;
- adding wild point to either regular or wild point results in a wild point.

Now, although you could write explicitly some wild points, nearly all of them are too hard to describe. The best approximation to the resulting picture is probably this: imagine a set of wild points $W$; the whole space is $W \times \mathbb Z^+\times \mathbb Z^+\times\cdots$. The remaining question, therefore, is what monoid $W$ is equivalent to. For that, you need to settle these questions:

- is it true that you cannot have $w_1 \oplus w_2 \oplus \cdots\oplus w_n = r$ where $r$ is a regular point;
- whether you can subtract them;
- whether and how wild points are divisible by naturals.

## Operations on wild points

I think now is certainly time to post another question :)