@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.