I've pondered the following problem without any real progress.
Show that $[0,2)\cdot_o\mathbb{N}=_o\mathbb{N}\neq_o\mathbb{N}\cdot_o[0,2)$.
I know that $P\cdot_o[0,2)=_oP+_oP$ follows by $f(a,b)=(b,a)$ as $$P\cdot_o[0,2)=(P\times\{0\})\cup(P\times\{1\})$$ and $$P+_oP=(\{0\}\times P)\cup(\{1\}\times P),$$ so my intuition tells me that this is usable to show that $$[0,2)\cdot_o\mathbb{N}=_o\mathbb{N}=_o\mathbb{N}\cdot_o[0,2)$$ as $$[0,2)\cdot_o\mathbb{N}=(\{0\}\times\mathbb{N})\cup(\{1\}\times\mathbb{N})$$ and $$\mathbb{N}\cdot_o[0,2)=(\mathbb{N}\times\{0\})\cup(\mathbb{N}\times\{1\})$$ which contradicts the result of the exercise.
I suppose that I could find some function which is order preserving to show the desired result, but I'm currently not making any progress towards such a function. I know that one can use that $\mathbb{N}$ lacks a maximum element to show that $$\mathbb{N}\neq_o\mathbb{N}+_o[0,1)$$ (as seen here) but I don't know what to do in my context (and I don't see why $[0,2)\cdot_o\mathbb{N}=_o\mathbb{N}$ works but $\mathbb{N}\cdot_o[0,2)\neq_o\mathbb{N}$).
When $<_A$ is a linear order on $A$ and $<_B$ is a linear order on $B$ we define the linear order $<_{A,B}$ as the reverse-lexicograpic order on $A\times B$: For $(a,b)$ and $(a',b')$ in $A\times B$ let $$(a,b)<_{A,B} (a',b') \iff ((b<_Bb') \lor (b=b' \land a<_Aa')).$$
When $A=\{0,1\}$ and $B=\mathbb N,$ with the usual orders on $A$ and $B$, the members of $A\times B$ with the order $<_{A,B}$ can be listed in increasing order: $$(0,1),(1,1),(0,2),(1,2),(0,3),(1,3),...$$ which is order-isomorphic to $\mathbb N.$
The members of $B\times A$ with the order $<_{B,A}$ can be listed as 2 increasing sequences: $$(0,1),(0,2),(0,2),...$$ followed by $$(1,1),(1,2),(1,3),...,$$ with $x<_{B,A}y$ whenever $x$ belongs to the first sequence and $y$ belongs to the second. This is NOT order-isomorphic to $\mathbb N,$ as $(1,1)$ has infinitely many predecessors and infinitely many followers, a property not seen in $\mathbb N.$
Think of $\{0,1\}\times \mathbb N$ as "$\mathbb N$" consecutive copies of $\{0,1\},$ and $\mathbb N \times \{0,1\}$ as $2$ consecutive copies of $\mathbb N.$