I am trying to classify such fibrations $p:E\to B$. I understand how to build a cochain representing the obstruction to building a section, and showing it is a cocycle, getting a cohomology class in $H^{n+1}(B;A)$. I want to show any other fibration with the same obstruction class is homotopic.
I know abstractly that the obstruction class corresponds to the homotopy class of some map $f:B\to K(A,n+1)$. If my original fibration was also induced by a map, I think I could show it must be in the same homotopy class, but I cannot convince myself that every fibration is actually induced, i.e., that the path fibration over $K(A,n+1)$ is `universal'.
Can someone give me a nudge in the right direction?
For a principal fibration with fiber $G$ to be universal, it suffices that its total space is (weakly) contractible. In your situation, the path fibration $$G = K(A, n) \to PK(A,n+1) \to K(A,n+1)$$ has total space $PK(A,n+1)$, which is contractible.
Let $$G \to EG \to BG$$ be a principal $G$-fibration with contractible total space. To show contractibility of the total space is a sufficient condition to induce all other principal bundles via pullback, suppose we are given another principal fibration $p: E \to B$ with fiber $G$. Consider an "intermediate" principal $G$-fibration $$E \times_G EG \to E \times_G \{*\} \simeq B.$$ The fiber of this fibration is equivalent to $EG$, which is contractible. Hence there exists a section $B \to E \times_G EG$. The classifying map $B \to BG$ is obtained by the composition $$B \to E \times_G EG \to \{*\} \times_G EG \simeq BG.$$
Check that the pullback of $EG \to BG$ along this map gives a fibration isomorphic to the original fibration $E \to G$.