Trouble with operator-sum representation of a quantum operation

You are completely right with the first part of your question: What is meant by $$|b_i\rangle_B$$ is indeed $$\mathbb I_A\otimes |b_i\rangle_B$$. (Note that omitting identities is quite common, e.g., when writing Hamiltonians!)

Now to your question how to show $$$$\begin{split} \sum_{i}{\langle b_{i}|\big(U(\rho_{A}\otimes |0\rangle\langle 0|)U^{\dagger}\big)|b_{i}\rangle}\overset{*}{=}\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}\ . \end{split}$$$$ To this end, note that \begin{align} \rho_{A}\otimes |0\rangle\langle 0| &= (\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle)(\mathbb I\otimes \langle 0|) \\ &= (\mathbb I\otimes|0\rangle) \rho_A (\mathbb I\otimes \langle 0|)\ . \end{align} (I elaborate below why this equality holds.) Inserting this on the LHS, we obtain $$$$\begin{split} \mbox{LHS}=\sum_{i}{\langle b_{i}|U (\mathbb I\otimes|0\rangle) \rho_A (\mathbb I\otimes \langle 0|) U^{\dagger}|b_{i}\rangle} =\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}\ , \end{split}$$$$ as desired.

Appendix: Why is $$(\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle)= (\mathbb I\otimes|0\rangle) \rho_A$$?

First, note that $$(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$$. Also note that we can regard $$|0\rangle$$ as a ($$d\times 1$$) matrix and $$1$$ as a ($$1 \times 1$$) matrix. We thus have \begin{align} (\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle) &= (\rho_A\mathbb I)\otimes(\mathbb I|0\rangle) \\ &= (\mathbb I\rho_A)\otimes(|0\rangle\cdot 1) \\ &= (\mathbb I\otimes |0\rangle)(\rho_A\otimes 1) \\&= (\mathbb I\otimes|0\rangle) \rho_A\ . \end{align}