After resubstitution you get:
$((A \lor B) \land \neg A) \lor ((A \lor B) \land B) $
Well, that's more complicated, yes, but still not too hard. For example, we can now apply Distribution on the left term:
$(A \lor B) \land \neg A) \Leftrightarrow (A \land \neg A) \lor (B \land \neg A) \Leftrightarrow \bot \lor (B \land \neg A) \Leftrightarrow B \land \neg A$
Plugging that back in, we get:
$(B \land \neg A) \lor ((A \lor B) \land B) $
Now, what about the right side? If you apply Distribuition here, you get:
$(A \lor B) \land B \Leftrightarrow (A \land B) \lor (B \land B) \Leftrightarrow (A \land B) \lor B$
... hmmm, that doesn't look too promising ... in fact, Distribution once more gets you to:
$(A \land B) \lor B \Leftrightarrow (A \lor B) \land (B \lor B) \Leftrightarrow (A \lor B) \land B$
... and we're back at where we were! ... OK, so that doesn't work.
Fortunately, here is something we can do:
$(A \lor B) \land B \Leftrightarrow (A \lor B) \land (\bot \lor B) \Leftrightarrow (A \land \bot) \lor B \Leftrightarrow \bot \lor B \Leftrightarrow B$
Aha! So, plugging that back in, we get:
$(B \land \neg A) \lor B$
and now we can do a similar trick:
$(B \land \neg A) \lor B \Leftrightarrow (B \land \neg A) \lor (B \land \top) \Leftrightarrow B \land (\neg A \lor \top) \Leftrightarrow B \land \top \Leftrightarrow B$
Now, the 'tricks' we did at the end actually reveal an important logical equivalence, known as:
Absorption
$P \land (P \lor Q) \Leftrightarrow P$
$P \lor (P \land Q) \Leftrightarrow P$
Also, the transformation at the beginning is known as:
Reduction
$P \land (\neg P \lor Q) \Leftrightarrow P \land Q$
$P \lor (\neg P \land Q) \Leftrightarrow P \lor Q$
So, with those, let's do the whole transformation again:
$((A \lor B) \land \neg A) \lor ((A \lor B) \land B) \overset{Reduction, Absorption}{\Leftrightarrow} (B \land \neg A) \lor B \overset{Absorption}{\Leftrightarrow} B$
Finally, though, I want to point out that the Distribution equivalence works both ways (it's an equivalence, remember?), so you could have done a 'Reverse' Distribution on your very original expression:
$(A \lor B) \land (\neg A \lor B) \overset{Distribution}{\Leftrightarrow} (A \land \neg A) \lor B \Leftrightarrow \bot \lor B \Leftrightarrow B$
And that equivalence actually has its own name as well:
Adjacency
$(P \lor Q) \land (P \lor \neg Q) \Leftrightarrow P$
$(P \land Q) \lor (P \land \neg Q) \Leftrightarrow P$
And with that, we can rework your transformation one more time:
$(A \vee B) \wedge (\neg A \vee B) \overset{Adjacency}{\Leftrightarrow}B$
Done!
But here's the moral: Put Absorption, Reduction, and Adjacency immediately into your 'Boolean Algebra Tool Box': they are very useful!