ct.category theory - From tensor algebras in monoidal categories to (commutative?) monoids

EDIT: As Peter points out in the comments, my answer to number 2 is incorrect. When $D$ is braided, you can "pull out the twist" using naturality of the braiding, but for general monoidal categories, this morphism will not be equal to $beta otimes alpha$.

However, there is an alternative procedure to the one you describe that always produces $beta otimes alpha$, as shown in this even more crudely drawn diagram:

This map corresponds to tensoring by $L^{otimes - j}$ in the middle rather than on the right. In either case, $alpha otimes beta neq beta otimes alpha$.

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1. Yes, by coherence. For instance, if you take a strict monoidal category $D'$ equivalent to $D$, then any bracketing of $L^{otimes i}$ in $D$ identifies with the canonical $L^{otimes i}$ in $D'$ in a manner compatible with taking homs and tensor products. This guarantees that all ways you choose to bracket your $L^{otimes i}$ and apply associators produce the same result, so your monoid is well-defined.
   
2. The result of your manipulations is $beta otimes alpha$. This can be checked most easily using string diagrams. Here is my rather crude attempt to illustrate this:
   
   
   
   The $beta$ can be "pulled around" to the left side of $alpha$. However, this morphism is not the same as $alpha otimes beta$ in general: this diagram is not isotopic to the one for $alpha otimes beta$, so the result fails in the "free" case.
   

Note that the answers to both parts of your question are forms of coherence. The first is perhaps most easily seen as a case of "every monoidal category is equivalent to a strict one," while the second is most easily seen as a case of "isotopic string diagrams represent the same morphism."