How is the concept of $\Bbb Z/2\Bbb Z$ extended to $\Bbb N/2\Bbb N$? Is $\Bbb N/2\Bbb N$ ambiguous? 
How is the concept of $\Bbb Z/2\Bbb Z$ extended to $\Bbb N/2\Bbb N$?  Is $\Bbb N/2\Bbb N$ ambiguous?

I believe $\Bbb Z/2\Bbb Z$ to be unambiguous terminology for cosets of the form $\{0,1\}+2\Bbb Z$ which is a representation of the cyclic abelian group $C_2$ having 2 elements and satisfying
$0+0=0, \\0+1=1, \\1+1=0.$
The difference between the group and the monoid quotient arises because in the case of $\Bbb N$, the set of additive transformations between two elements of the monoid, is NOT the monoid $\Bbb N$, it is the group $\Bbb Z$.
Therefore there are two ways to extend this idea to a quotient on the monoid $\Bbb N$.
Way one, which seems most natural to me is to define equivalence classes using the GROUP $\Bbb Z=\Bbb N-\Bbb N$ as follows: $x\sim y\iff\exists i\in\Bbb Z:x-y=2i$
Then we have $\Bbb N/2\Bbb N\cong\{0,1\}+2\Bbb N$ which is, heretofore, my meaning of a monoid quotient.  But, is this a normal interpretation of $\Bbb N/2\Bbb N$?

Way two, if one was minded to do so, would be to argue the claim that $1+3\neq0$ because there does not exist $-4\in\Bbb N:4-4=0$.  Then we would have, well, something relatively degenerate but off the top of my head we'd require left- and right- equality, and equality would be a kind of equivalent mod 2 and greater than, or equivalent mod 2 and less than.

Is the first case not the accepted meaning of a monoid quotient?

Here is the problem I have:  Whenever I do precisely the same as way one above, but in a multiplicative monoid, as I do here I receive the objection analogous to the statement that $\Bbb N/2\Bbb N\neq\{0,1\}+2\Bbb N$ in this additive monoid.  So I need to clear up whether it is I using unclear or just plain wrong terminology.
 A: I guess you real question is :let $M$ be a commutative monoid, written additively, and let $N$ be a submonoid. How to define properly a quotient monoid $M/N$ ?
First, we need a good equivalence relation on $M$ (and this is why you got into trouble. You didn't have the right one). Say that $m\sim_N m'$ if there 
exists $n,n'\in N$ such that $m+n=m'+n'$.
This is indeed an equivalence relation on $M$, because we have a commutative monoid $M$, and $N$ is a submonoid (I leave you the details).
Call $M/N$ the quotient set. Notice that if $m_1\sim_N m'_1$ and $m_2\sim_N m'_2$, then $m_1+m'_1\sim_N m_2+m'_2$ (once again because we have a commutative monoid and $N$ is a submonoid), so you get a law on $M/N$ defined by $\overline{m}_1+\overline{m}_2=\overline{m_1+m_2}$, which makes $M/N$ a monoid, such that the canonical map $M\to M/N$ is a morphism of monoid.
If you apply all of this to $M=\mathbb{N}$ and $N=2\mathbb{N}$, you will get what you want.
A: In the field of universal algebra there are congruences and that is (I think) what you are looking for. A congruence relation is basically an equivalence relation that takes into account the algebraic structure. Congruence relations correspond bijectively to quotient algebraic structures. Notion of a congruence can be thought of as a generalization of notions of a normal subgroup of a group and two-sided ideal of a ring etc. For instance consider a group $G$ and its normal subgroup $N$. Then 
$$x\equiv y \mbox{ iff }xy^{-1}\in N$$ 
is a congruence relation and every congruence relation on $G$ is of this form (i.e. it comes from a unique normal subgroup of $G$).
Let $\left(\mathbb{N},+,0\right)$ be a monoid. Consider a congruence relation $\equiv$ on this monoid such that one of the equivalence classes of $\equiv$ is precisely the set of even numbers $2\mathbb{N}$. Then your question can be rigorously stated as follows:

Is the congruence $\equiv$ unique?

The answer is yes. Pick two odd natural numbers $a, b\in \mathbb{N}\setminus 2\mathbb{N}$. Without loss of generality $a <b$. Then $b- a$ is even. Hence $b-a \equiv 0$. Now since $\equiv$ is a congruence, we derive that 
$$a + (b-a) \equiv a+0$$
Thus $a\equiv b$ and hence all odd numbers are equivalent. So $\equiv$ has only two equivalence classes $2\mathbb{N}$ and $1+2\mathbb{N}$.
Remark.
You can check that the equivalence relation $\equiv$ with equivalence classes $2\mathbb{N}$ and $1+2\mathbb{N}$ is a congruence for $\left(\mathbb{N},+,0\right)$.
