@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 !