I've been thinking about the case when \$$\otimes\$$ is not symmetric.

As @Matthew says, we can turn \$$\mathcal{V}\$$ into a \$$\mathcal{V}\$$-enriched preorder if \$$(x\otimes -)\$$ has a right adjoint.

If on the other hand it's \$$(-\otimes x)\$$ that has a right adjoint, the proof doesn't work. However in this case we could slightly vary the definition of a \$$\mathcal{V}\$$-enriched category to use right-to-left composition instead of left-to-right composition.

This amount to replacing the \$$\mathcal{V}(x, y)\otimes\mathcal{V}(y, z)\leq\mathcal{V}(x, z)\$$ condition with \$$\mathcal{V}(y, z)\otimes\mathcal{V}(x, y)\leq\mathcal{V}(x, z)\$$ – similarly to how we define \$$\text{hom}\$$ in a \$$\textbf{Set}\$$-category. Then the proof would go through.

So whether to go with \$$(x\otimes -)\$$ or \$$(-\otimes x)\$$ depends on which way round we write composites.

A speculative interpretation of this: we could read the non-symmetric \$$A \otimes B\$$ as "A and then B". In this case a right adjoint of \$$A \otimes -\$$ would be like a permit to turn any A *that you've already got* into a B, because \$$A \otimes (A \multimap B) \leq B\$$, while a right adjoint of \$$- \otimes A\$$ would be a permit to turn any A *that you get in the future* into a B, because \$$(A \multimap B)\otimes A \leq B\$$. Maybe there's some "time-conscious" logic sitting in here that might be useful in some situations.