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.