Let’s say I have a set $A$. I build the free commutative monoid $M$ generated by $A$.


1. When can a well-order on $A$ be extended to $M$, in a way that is compatible with its monoid structure? I am interested both in the case when $A$ is finite and infinite.
2. Under which conditions would such an extension be unique?



Generating totally ordered free commutative monoids
For your first question, the answer is positive with lexicographic ordering (as said by Chris). 
Your monoid is $\mathbb{N}^{(A)}$ (i.e. the set of mappings $\alpha: A\to \mathbb{N}$ with finite support, additive version) or $M=\{A^{\alpha}\}\_{\alpha\in \mathbb{N}^{(A)}}$ for a multiplicative version using the [multiindex notation](https://en.wikipedia.org/wiki/Multi-index_notation) like in the multivariate polynomials (take e.g. the ring $\mathbb{Z}[A]$).


An easy way to see that this is an ordering is to consider the free commutative group over the same alphabet $A$, i.e. $\mathbb{Z}^{(A)}$ and ($A$ being still supposed totally ordered by 
$<$) the set 
$$
P:=\{\alpha\in \mathbb{Z}^{(A)}\setminus\{0\}\ |\ \alpha(\min(supp(\alpha)))>0\}
$$ 
where the support of $\alpha$ is
$$
supp(\alpha)=\{a\in A\ |\ \alpha(a)\not=0\}
$$
in other words $P$ is just the set of non-zero multiindices such that the coefficient at the least place is $>0$ (for this only total order is required). One has just to check that 


1. $P+P\subset P$- $P\cap -P=\emptyset$- for $\alpha,\beta\in \mathbb{N}^{(A)}$, $\alpha\prec\_{lex}\beta$ iff 
$\beta-\alpha\in P$- $\prec\_{lex}$ extends the given ordering on $A$- if $(A,<)$ is well ordered, so is $(\mathbb{N}^{(A)},\prec\_{lex})$


which is maybe long but straightforward. 


For your second question, one can see that this extension is unique iff $|A|\leq 1$ ($A$ is empty or a singleton) as, if $|A|\geq 2$ all lexicographic orderings are different 
(and exchanged by permutations in case $A$ is finite). Still if $|A|\geq 2$ (and totally ordered), there are orders which are not in this class as the "graded lexicographic ordering 
$\prec\_{grlex}$" obtained by 
$$
\alpha\prec\_{grlex}\beta \Longleftrightarrow |\alpha|<|\beta|\mbox{ or }(|\alpha|=|\beta| 
\mbox{ and } \alpha\prec\_{lex}\beta)
$$ 
... and in all cases (when $A$ is not empty), the reverse ordering is different from the given one.


Hope it helps, do not hesitate if something is unclear.


