The final bit of the puzzle is proving the second triangle equation from the first one and the pentagon. I'll just sketch a proof here.

The trick here is to tensor the second triangle equation by \$$A\$$ on the left. This looks like it's making it more complicated, but the fact we now have a four-fold product allows us to use the pentagon diagram to go from \$$A\otimes ((I\otimes B$$\otimes C)\\) to \$$A\otimes (I\otimes (B\otimes C))\$$ the "long way round".

At each stage we knock out the \$$I\$$ with a suitable unitor. This gives us the following diagram:

The quads (2) and (4) are both naturality diagrams for associators. The triangles (3) and (5) are both instances of the first triangle equation (in the case of (3) we've tensored everything by \$$I\$$ on the right). Note also that \$$1_{B\otimes C} = 1_B \otimes 1_C\$$, which is what allows us to "glue" the sides of (4) and (5) together.

Now since the outer pentagon commutes, and (2), (3), (4), (5) commute, we must have (1) commutes. To remove the \$$A\$$, we simply set it equal to \$$I\$$ then apply the natural isomorphism \$$\lambda\$$ to get our second triangle equation \$$\lambda_B\otimes 1_C = \lambda_{B\otimes C}\circ\alpha\$$.