Definition of a monoid: clarification needed

First definition (algebraic): A monoid is a pair $(M,b)$, where $M$ is a set (called the underlying set of the monoid) and $b\colon M\times M\to M$ is a mapping (called the binary operation of the monoid; for $m_1,m_2\in M$ denote $b(m_1,m_2)=m_1\bullet m_2$), which satisfy the following two properties:

1). for any $m_1,m_2,m_3\in M$ the following equality holds (associativity): $$(m_1\bullet m_2)\bullet m_3=m_1\bullet(m_2\bullet m_3).$$

2). there exists such $e\in M$ (called the identity of the monoid), that for any $m\in M$ the following equalities hold: $e\bullet m=m\bullet e=m$.

It is a standard definition of a monoid. There are a lot of examples of monoids; for example, any group is a monoid. However, there are monoids, which are not groups.

You are right, $(\mathbb{Z},+)$, $(\mathbb{Z},\cdot)$ are monoids, but $(\mathbb{Z},-)$ is not (because, for example, $1-(1-1)\ne(1-1)-1$).

If you are familiar with category theory, then you can get a lot of natural examples in different categories. The reason is following:

Let $A$ be a category, $a\in A$ be its object. Then any morphism $f\colon a\to a$ is called an endomorphism of $a$. Thus we can consider the set of all endomorphisms of the object $a$, denote it by $end(a)$. Note, that we can composite any two morphisms in $end(a)$, therefore we get a binary operation on $end(a)$! $$ \bullet\colon end(a)\times end(a)\to end(a);\qquad f\bullet g=f\circ g. $$ It is a monoid by the definition of category.

Now you can take any category, for example $\mathbf{Set}$, take any its object -- for example, $\mathbb{N}$ (which is considered without its algebraic structure), -- and you get the monoid $\mathbb{N}^{\mathbb{N}}$ of all functions $f\colon\mathbb{N}\to\mathbb{N}$, which, of course, is not a group (check it).

Now it is easy to believe in the following definition of a monoid:

Second definition (category-theoretic): Monoid is a category with one object.

Indeed, denote by $x$ its single object, then $end(x)$ is a corresponding monoid. Conversely, if you have a monoid $(M,b)$, you can define a category $\mathbf{M}$ with single object, such that $Arr(\mathbf{M})=M$, and composition defined by the following rule: $$ \forall f,g\in M:\;f\circ g=f\bullet g. $$

But it was, of course, intuitive reasoning. In order to make an exact statement, I have to give a few more definitions:

1). Let $(M,\bullet_M)$ and $(N,\bullet_N)$ be monoids. Monoid homomorphism is a mapping $f\colon M\to N$, such that for all $m_1,m_2\in M$ the equalities $f(m_1\bullet_M m_2)=f(m_1)\bullet_N f(m_2)$ and $f(e_M)=e_N$ hold. It's easy to check that all small monoids and their homomorphisms form a category, called $\mathbf{Mon}$.

2). Let's denote by $\mathbf{S}$ the full subcategory of $\mathbf{Cat}$, which objects are categories with fixed single object.

Now I can formulate the exact statement (equivalence of two definitions):

Statement: The category $\mathbf{Mon}$ of monoids is isomorphic to the category $\mathbf{S}$.

For your first question, if you compare the axioms of a monoid and a category, you will see, they are quite similar (associativity and identities). So if you have a category with a single object, the endomorphism ring of this object will be a monoid. This also holds more generally: For each category, the endomorphism ring of an object is a monoid.

For examples of monoids: First, the ones that are already present in your explanation, e.g. $(\mathbb{R},+)$, $(\mathbb{Q},+)$, $(\mathbb{R},\cdot)$, $(\mathbb{R}\setminus \{0\},\cdot)$, $(\mathbb{N},+)$, etc.

Another, maybe more exotic example for a high school student is the free monoid. It is given by words in some alphabet. An alphabet is just some set, e.g. $\{a,b\}$. In this examples words include e.g. $abaab$ or $baaabbba$. The binary operation is given by concatenating two words, e.g. $abaab\cdot baaabbba=abaabbaaabbba$. The identity element is here given by the empty word.

Or, if you take the answer to your first question, and plug in as an example the category of sets and the object being the real numbers. You get the set of functions from $\mathbb{R}$ to $\mathbb{R}$ form a monoid.