(Here is an nlab link talking about what I describe below)

It's too cumbersome to ask for all possible ways to interpret a product to be identical, which is why we merely require there to be *coherence isomorphisms* — a natural isomorphism between any two ways to multiply out a product. This even including inserting and removing copies of the empty product!

The unitors $A \otimes I \cong A$ and $I \otimes A \cong A$ and the associator $(A \otimes B) \otimes C \cong A \otimes (B \otimes C)$ are examples of this. But we also expect, for example, there to be a coherence isomorphism

$$ (A \otimes (B \otimes I)) \otimes (C \otimes I) \cong (A \otimes (B \otimes C)) \otimes (I \otimes I)$$

The important thing, however, is that we *can* (and *should*) demand that coherence be unique — that any way of combining these coherence isomorphisms gives the canonical one.

For example, I could build another isomorphism between the two products via:

$$ (A \otimes (B \otimes I)) \otimes (C \otimes I) \cong
(A \otimes B) \otimes C
\cong (A \otimes (B \otimes C)) \otimes (I \otimes I)$$

Composing these two isomorphisms is guaranteed to be the same isomorphism as the one in the previous equation. The same is true if I construct an isomorphism out of the associator and unitor natural isomorphisms.

The pentagon is the smallest example of how to use the associator isomorphism to construct two distinct paths between interpretations of the product; the pentagon identity is the one asserting that the two paths give the same isomorphism.

I think it's clear that the unitors and the associator are enough to produce a coherence identity between any two ways of multiplying out a product. It's maybe somewhat surprising, though, that all of the coherence identities they have to satisfy are generated by the pentagon and the triangle (which relates the unitors and the associator).