> Is that right?

It's certainly a monoid, but I don't think it respects the partial order structure. Since \\(c \le a\\) and \\(d \le d\\), we expect \\(c \otimes d \le a \otimes d\\). But \\(c \otimes d = c \not\le d = a \otimes d\\).

Here's a multiplication table I built out of your rule:

|abcd
-+----
a|abcd
b|bccc
c|cccc
d|dccc