@Jonathan – Ah okay! and you can simplify this to give the counterexample I'm looking for:

Take the trivial monoid \\(\mathbf{1}\\) and the 2-element join semilattice \\((\mathbf{2}, \vee, \bot)\\)

Let \\(f : \mathbf{1} \rightarrow \mathbf{2}\\) be the map that picks out \\(\top\\)

Then \\(f(\bullet) \otimes f(\bullet) = f(\bullet) \vee f(\bullet) = \top \vee \top = \top = f(\bullet \otimes \bullet)\\)

So \\(f\\) preserves \\(\otimes\\). But \\(f(\bullet)\neq\bot\\), so \\(f\\) does not preserve identity.