What are the domains of the multiplication and unit morphisms of a monoid object? I'm trying to understand what a Monoid is from category theory perspective, but I'm a bit confused with notation used to describe it. Here is Wikipedia:

In category theory, a monoid (or monoid object) $(M, \mu, \eta)$ in a monoidal category $(\mathcal{C}, \otimes, I)$ is an object $M$ together with two morphisms
  
  
*
  
*$\mu: M \otimes M \to M$ called multiplication,
  
*$\eta: I \to M$ called unit, [...]
  

My confusion is about morphism notation. Why is the binary operation $\otimes$ a part of the morphism notation? My understanding of morphism is that it's a kind of function that can map one type to another (domain to codomain) like $M \to M$... Why is the operation $\otimes$ a part of the domain in the definition?
The second confusion is about $I$. Why $I$ is a domain... there is no $I$ object in a Monoid at all. It's just the neutral element of object $M$.
I understand that a Monoid is a category with one object, identity morphism and binary operation defined on this object, but the notation makes me think that I don't understand something.
Is $M \otimes M$ somehow related to cartesian product, so the domain of the morphism is defined as $M \times M$ ?
 A: An ordinary monoid is a set $M$ equipped with a map $\mu\colon M\times M\to M$ and an element $e\in M$ such that certain axioms are satisfied. 
Now the goal of the definition you quoted (monoid object in a monoidal category) is to generalize the notion of ordinary monoid to the most general categorical context possible. 
The first step is to observe that in a general category it doesn't make any sense to talk about an "element" of an object. So let's replace the element $e\in M$ with a map $e\colon 1\to M$, where $1 = \{*\}$ is a set with a single element. The point is that (in the category of sets) specifying a map $1\to M$ is exactly the same as specifying an element of $M$: given an element $e\in M$, we can define a map $1\to M$ by $*\mapsto e$, and given a map $e\colon 1\to M$, we can define an element of $M$ by $e(*)$. 
Ok, now we want to generalize away from the category of sets and talk about monoids living in other categories. So let $\mathcal{C}$ be a category. A monoid in $\mathcal{C}$ should be an object $M$ in $\mathcal{C}$ together with maps $\mu$ and $e$ satisfying certain axioms. To make sense of the maps $\mu$ and $e$, we need:


*

*A specified way of putting two copies of $M$ together to make the domain of $\mu$, analogous to $M\times M$ in the case of sets. Let's denote this by $M\otimes M$.

*A special object to be the domain of $e$, analogous to $1$ in the case of sets. Let's denote this by $I$. 


The operation $\otimes$ on objects and the special object $I$ are the basic data of a monoidal category. 
If $\mathcal{C}$ has finite products, you're welcome to define $\otimes$ to be the product $\times$ and $I$ to be the terminal object $1$. This is called the Cartesian monoidal structure. But the point is that we don't need $M\otimes M$ to be the product and $I$ to be the terminal object in order to make sense of the definition of monoid. 
We do need to assume a little bit more about $\otimes$ and $I$ in order to make sense of the monoid axioms. In particular, we need $\otimes$ to be associative (up to isomorphism), $$M\otimes(M\otimes M) \cong (M\otimes M)\otimes M,$$ and we need $I$ to act as an identity for $\otimes$ up to isomorphism, i.e. $$M\cong I\otimes M\cong M\otimes I.$$
Making all of this appropriately functorial and natural leads directly to the full definition of monoidal category. 
