In algebra, a monoid is defined as a pair $(M, *)$ where $M$ is a set and $*$ is a binary operation on $M$ satisfying associativity and having an identity element.
A category with a single object, on the other hand, is given by a triple $(O, M, *)$ where $O$ is a set with a single element (the class of objects), $M$ is a set (the set of morphisms from the single object to itself is the only hom-set) and $*$ is a binary operation on $M$ satisfying associativity and having an identity element (the composition of morphisms turns out to be a binary operation because there is only one hom-set).
Notice that in the definition of a category with a single object there is no relationship between $O$ and $(M, *)$, because the properties that need to hold involve only $(M, *)$. Since those properties are exactly the same as those in the definition of monoid, we can say that a category with a single object consists of a monoid together with a set with a single element.
Therefore, if you want to get a monoid from a category with a single element, you just have to discard the useless set with a single element. Conversely, if you want to get a category with a single element from a monoid, you just have to come up with a set with a single element. Since that element won't play any role in the category, one usually puts a placeholder like $@$ to mean that any object will do.
Two categories with a single element are isomorphic (not just equivalent) whenever the corresponding monoids are isomorphic. Indeed, if $f\colon M \to M'$ is a monoid isomorphism, then the functor $F \colon \mathcal C \to \mathcal C'$ defined by $F(@)=@'$ and $F(m) = f(m)$ for any $m \in M$ (that is $m \colon @ \to @$ as a morphism) is clearly an isomorphism of categories, its inverse $F^{-1} \colon \mathcal C' \to \mathcal C$ being given by $F^{-1}(@') = @$ and $F^{-1}(m')=f^{-1}(m')$ for any $m' \in M'$ (i.e., $m' \colon @' \to @'$).
Now, if you still think that it makes more sense to define a monoid as a category with a single object, and that in order for a monoid defined in this way to be interesting its object should be some algebraic object as well, then you could consider the monoid of endomorphisms of a given algebraic object.
For example, if $(R, +, *)$ is a ring, then you could consider the full subcategory of the category of rings having $(R, +, *)$ as its only object. In this way there is indeed a relationship between the object and the morphisms, since the morphisms are precisely the ring homomorphism from $(R, +, *)$ to itself.
In fact, this construction can be applied to any category, not just a category of algebraic objects (e.g., you could also consider the full subcategory of the category of topological spaces having a given topological space as its only object, and so on).
The only object in a monoidal category with one object is the unit $I$. Thus the only morphisms are those from $I$ to itself. These include the identity morphism and are closed under the associative composition, so form a monoid by definition.
Now the monoidal structure necessitates that $I\otimes I$ is the same as $I$. Consequently, given two morphisms $f,g\colon I\to I$, you can interpret $f\otimes g\colon I\otimes I\to I\otimes I$ as a morphism $I\to I$. Thus, $\otimes$ gives a second way of combining morphisms $I\to I$. Moreover, the properties of $\otimes$ require that identity morphisms are a unit for this way of combining as well, and that it distributes over the first way: $ff'\otimes gg'=(f\otimes g)(f'\otimes g')$. This is the "monoid in the category of monoids" that is meant.
Best Answer
I think you are confusing the universal property of the initial object of a category and the universal property of a free object.
What follows may be a bit too abstract for your taste, so you can skim and skip to the portion below the horizontal line:
Definition. Given a category $\mathcal{C}$, an initial object in $\mathcal{C}$ is an object $s$ such that for each object $a\in \mathcal{C}$ there is exactly one arrow $s\to a$.
An initial object, if it exists, is unique up to unique isomorphism: if $s$ and $s'$ are both initial objects, then there is a unique map $f\colon s\to s'$, a unique map $g\colon s'\to s$. So $g\circ f\colon s\to s$ is the unique map from $s$ to $s$, so $g\circ f = \mathrm{id}_s$; and similarly $f\circ g = \mathrm{id}_{s'}$.
By contrast, a free object is defined in terms of sets:
Definition. Let $\mathcal{C}$ be a category in which every object is a set and every arrow is a set function. Let $\mathbf{U}\colon\mathcal{C}\to\mathcal{S}et$ be the underlying set functor that maps every object to its underlying set. If $X$ is a set, then a free $\mathcal{C}$-object on $X$ is an object $F(X)$ of $\mathcal{C}$, together with a function $i\colon X\to \mathbf{U}(F(X))$ from $X$ to the underlying set of $F(X)$, such that:
A free object on $X$, if it exists, is unique up to unique isomorphism. But the map $f$ need not be onto.
If $\mathcal{C}$ has free objects on every set, then the free object on the empty set is necessarily the initial object in $\mathcal{C}$. Thus, the trivial group is the initial object in $\mathcal{G}roup$, and the trivial monoid is the initial object in $\mathcal{M}onoid$, because the trivial group (resp. monoid) is the free group (resp. monoid) on the empty set. On the other hand, there is no free semigroup on the empty set.
The free $\mathcal{C}$-object on $X$, when it exists, can be interpreted as an initial object, but in a different category: you consider the category of all pairs $(C,j)$, where $C$ is an object of $\mathcal{C}$, $j\colon X\to C$ is a set-theoretic function, and arrows $(C,j)\to (D,k)$ are morphisms $f\colon C\to D$ from $\mathcal{C}$ such that $f\circ j = k$. Then $(F(x),i)$ is an initial object of this category.
If $\Sigma^*$ is the monoid of all strings drawn from the alphabet $\Sigma$, then it turns out that $\Sigma^*$ is the free monoid on the set $\Sigma$. The universal property it satisfies is not being "an initial object in the category of monoids", but rather that described above: given any set-theoretic function $j\colon \Sigma\to M$ into a monoid $M$, there is a unique monoid homomorphism $f\colon \Sigma^*\to M$ such that $f|_{\Sigma}=j$ (that is, $f$, restricted to the set $\Sigma$ viewed as a subset of $\Sigma^*$, is equal to $j$). But this map is not necessarily onto. It is onto if and only if $j(\Sigma)$ generates $M$ (as a monoid).
The proof that $\Sigma^*$ is indeed the free monoid on $\Sigma$ is fairly straightforward. Given $j\colon \Sigma \to M$, we define $f\colon \Sigma^*\to M$ recursively, using the definition of $\Sigma^*$:
This defines $f$ on all of $\Sigma^*$. Using associativity, you show that this is a monoid homomorphism. And for uniqueness, you again do it by induction: any other $g$ which agrees with $f$ on $\Sigma_1$ must agree on $\Sigma_0$ (because they are module homomorphisms), and if they agree on $\Sigma_i$, then they will agree on $\Sigma_{i+1}$, since $g(uv) = g(u)j(v) = f(u)j(v)=f(uv)$.
This shows that for any monoid $M$ and any set-theoretic function $j\colon \Sigma\to M$, there is a unique monoid homomorphism $f\colon \Sigma^*\to M$ such that $f(\sigma)=j(\sigma)$ for all $\sigma\in\Sigma$, so $\Sigma^*$ is the free monoid on $\Sigma$.
Again, $f$ will be onto if and only if $\langle j(\Sigma)\rangle = M$. It is always onto $\langle j(\Sigma)\rangle$, since the image is a submonoid of $M$ that contains $\Sigma$, and every element in the image is a product of elements of $j(\Sigma)$.