@Anindya: that's awesome! (To use an A-word, as they're the best of all words.)

I like that your example is really simple at the core, since it's enough to work with only four of those words, let's say \$$A\$$, \$$B\$$, \$$C\$$ and the empty word; since you only care about the first letter anyway, we can define multiplication by just keeping the first letter, so that e.g. \$$C\otimes A = C\$$. This is much simpler than my own solution!

Here's my own solution. Take any poset \$$P\$$ with a nontrivial order; for example, \$$P = \\{0,1\\}\$$ with \$$0\leq 1\$$ will do just fine. Now consider the set of all maps \$$P \to P\$$. It's a monoid under composition: a function \$$f : P \to P\$$ and a function \$$g : P \to P \$$ multiply to \$$g\circ f : P \to P\$$, which is the function given by \$$(g\circ f)(x) := g(f(x))\$$. We can also equip this monoid with the pointwise order, which means that we write \$$f\leq g\$$ whenever \$$f(x) \leq g(x) \$$ in \$$P\$$ for all \$$x\$$. It's easy to see that if \$$f\leq f'\$$, then also \$$f\circ g \leq f'\circ g\$$ for any \$$g\$$. However, in general we cannot expect \$$g\leq g'\$$ to imply \$$f\circ g\leq f\circ g'\$$, since \$$f\$$ may even reverse some of the order relations! Concretely, choose some \$$y,y'\in P\$$ with \$$y\leq y'\$$ and \$$y'\not\leq y\$$, and take \$$g\$$ to be the constant function that maps every element to \$$y\$$, and let likewise \$$g'\$$ take everything to \$$y'\$$. And now choose any \$$f\$$ with \$$f(y) = y'\$$ and \$$f(y') = y\$$. VoilĂ !