Not every commutative semigroup is isomorphic with the multiplicative semigroup of a ring This is a statement in commutative algebra notes
page 19.
First, the multiplicative semigroup associated with the a ring $R$ should be $(R-\{0\}, \times)$, in my opinion, but the author claims that for a commutative semigroup to be the multiplicative semigroup of a ring it should contain an element acting like $0$. Why?
Second, in the example given, the author uses the addition operation on the commutative multiplicative semigroup of $\{x^k, k \in N\}$, but how do you define addition for formal monomials?
 A: The set $R\setminus \{0\}$ is not always closed under multiplication, so it would not be a semigroup. It is only closed under multiplication when the ring is a domain. The only reasonable definition of the multiplicative semigroup of a ring $(R,+,\cdot)$ is $(R, \cdot)$.
Any example of a commutative semigroup without an absorbing element is an example of a semigroup that can't be the multiplicative semigroup of a ring.
The set $\{x^n|n\in\mathbb N\}$ is such a semigroup, because there is no element which absorbs multiplication by the rest.
Actually the author improves upon this by showing that you can even have a zero element and not have a ring (but that would spoil the answer to the next part of the question, so stay tuned below.)

How do you define addition for formal monomials?

After a reading of the notes you are referencing, I have to say that this is the wrong question. There is no need to define addition for these. What the author is doing in that passage is taking $X=\{0\}\cup \{x^k\mid k\in\mathbb N\}$ and showing that if you suppose it has a ring structure (any ring structure at all) you arrive at a contradiction. The addition given to us by virtue of the supposition that "suppose $X$ is the multiplicative monoid of a ring."
