Turning Preordered Sets into Preordered Monoids (Constructing Preordered Monoids from Preordered Sets) Question: 
Referring to the Wikipedia article on Adjoint Functors in Section 2 (Motivation), they talk about "turning rngs into rings" (can be rephrased as "constructing rings from rngs").  I do not quite understand how this "turning" of algebraic structures into new algebraic structures is done. Which brings up my main question: knowing this method of turning structures into new structures, can we do the same with turning preordered sets into preordered monoids? Can we do the same with turning either structure into preordered free monoids?
Background: 
Definition (Preordered Monoid). A quadruple $\mathcal{X}=(X,\sqsubseteq,\cdot,e)$ such that


*

*$\sqsubseteq$ preorders $X$,

*$(X,\cdot,e)$ is a monoid,

*for all $x\in X, \ e\sqsubseteq x$, and

*(for all $w,x,y,z\in X$) if $w\sqsubseteq x$ and $y\sqsubseteq x$, then $w\cdot y\sqsubseteq x\cdot z$.


In general it is not possible to extend a given preorder $(X, \sqsubseteq)$ to a preordered monoid $(X, \sqsubseteq, \cdot, e)$:
Consider any preorder $(X, \sqsubseteq)$ such that for all $x \in X$ there is some $y \in X$ with $y \sqsubseteq x$ and $x \not \sqsubseteq y$. [For example $(\mathbb Z, \le)$]. Then there is no $e \in X$ such that $e \sqsubseteq x$ for all $x \in X$ and therefore, $(X, \sqsubseteq)$ cannot be extended to a preordered monoid $(X, \sqsubseteq, \cdot, e)$ without changing $\sqsubseteq$.
On the other hand, if $(X, \sqsubseteq)$ contains an element $e \in X$ such that $e \sqsubseteq x$ for all $x \in X$, we may let $\cdot \colon X \times X \to X, x \mapsto e$ be the trivial multiplication which yields a preordered monoid $(X, \sqsubseteq, \cdot, e)$ as desired.
I'm not sure what you mean by a free preorderd monoid in this setting, but if you provide your definition, I might be able to add an answer to that last part of your question as well.
