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.