Is $S$ a monoid, or is $(S,*)$ a monoid? If I have a set $S$ with operation $*$ as a monoid. Would I say I have a monoid $S$ with the binary operation $*$ or would I say I have a monoid $(S,*)$ where the binary operation $*$ does blahblahblah?

My book referred to $S$ as a monoid, which is strange considering $S$ is a set, and needs that operation to have certain qualities to be a monoid, so $S$ to me isn't at all a monoid...?

$(S,*)$ can be a group, but $S$ can't be, so the notation he used doesn't even seem consistent.
 A: Sometimes you see this:
$$
\mathfrak S = (S, *)
$$
and we say $\mathfrak S$ is a monoid, while $S$ is the carrier of $\mathfrak S$.
But, except in cases where we need to distinguish different monoids with the same carrier, it is much more common to use the same letter $S$ for both purposes.
A: I believe it's a matter of choice (and formality), both ways are right, given that the operation in the set is obvious from the context. Just note that if you already defined what a monoid is, in the second way you'd only need to say that $(S,*)$ is a monoid. 
What you'd need to do sometimes is explicitly define the operation saying something like "Let $(S,*)$ be a monoid with its operation defined as ..."
A: The same terminology is common for groups:

Let $G = \mathbb{R}^+$ under multiplication. Then $G$ is a group.

This is lazy writing, because formally we should not refer to any set as being a group, but only a set under some operation.  For instance we could say that $G$ is a group under the operation $*$.
But this usage is so common, and its meaning so obvious, that I think it is acceptable shorthand for the slightly more clunky statement "then $(G, *)$ is a group."
All of this applies equally well to monoids.
A: A set can't have an algebraic structure unless you specify the binary operation given to it.So now you see what you should say
But in some cases authors neglect saying redundant facts.So once $(S,\circ)$ is assumed to be monoid they reduce it to only $S$ is a monoid
