OFB output jth-block expression clarification
Is he implicitly saying that
[Cj-1 ⊕Pj-1] = Oj-1

Yes, this is true.

? If yes, why?

Well, we know that:
$$C_{j-1} = P_{j-1} \oplus E(K, O_{j-2})$$
(this is the first formula, replacing $j$ with $j-1$), and we know that:
$$O_{j-1} = E(K, O_{j-2})$$
Combining the two, we get:
$$C_{j-1} = P_{j-1} \oplus O_{j-1}$$
or, in other words:
$$C_{j-1} \oplus P_{j-1} = O_{j-1}$$