Hmm. I am curious: (this gets a bit stream of consciousness at points.)
Conjecture: Given \$$\mathcal{V}\$$ a monoidal poset, the following are equivalent:
\$i) \forall X\, Y\, \mathcal{V}-functors, \mathcal{X}(x,x') \otimes \mathcal{Y}(y,y') \otimes \mathcal{X}(x',x'') \otimes \mathcal{Y}(y',y'') \le \mathcal{X}(x,x'') \otimes \mathcal{Y}(y,y'') . \qquad (\star)\$
\$ii) \mathcal{V} \, \text{is commutative}\$

Proof attempt:

The proof for \$$ii \to i\$$ is in the OP.

For \$$i \to ii \\$$, let's try the contrapositive. If \$$\mathcal{V}\$$ is non-commutative then: \$\exists v_x v_y, v_x \otimes v_y \neq v_x \otimes v_y \$

So can we use that to construct a pair of \$$\mathcal{V} \$$-categories X and Y such that star is false?
\$\mathcal{X}(x,x') \otimes \mathcal{Y}(y,y') \otimes \mathcal{X}(x',x'') \otimes \mathcal{Y}(y',y'') \le \mathcal{X}(x,x'') \otimes \mathcal{Y}(y,y'') . \qquad (\star)\$
We can't use single object \$$\mathcal{V} \$$-categories, because \$$v_0\$$, \$$v_1\$$ might be less then \$$I\$$, but if we have exactly two objects, then Hom(0,1) can be whatever we want, with the other hom-objects being \$$I \$$. Let \$$\mathcal{X} \$$ be such with \$$Hom(0,1) = v_x\$$ and Y the same with \$$v_y\$$.

Ohh, interesting, if \$$v_x \otimes v_y \equiv v_y \otimes v_x\$$ then star is is still always true.
So we need to change \$ii) \mathcal{V} \, \text{is commutative}\$ to \$ii) \mathcal{V} \, \text{is symmetric (or braided? Not sure what those terms mean exactly)}\$. (

and change \$\exists v_x v_y, v_x \otimes v_y \neq v_y \otimes v_x \$ to \$\exists v_x v_y, v_x \otimes v_y \nleq v_y \otimes v_x \$

and now [\mathcal{X}(0,1) \otimes \mathcal{Y}(0,0) \otimes
\mathcal{X}(1,1) \otimes \mathcal{Y}(0,1) = v_x \otimes I \otimes I \otimes vy = v_x \otimes v_y

Wait the identities means that that does equal what its supposed to..uh..I think I am lost.