A* graph search time-complexity
These are basically two different perspectives or two different ways of viewing the running time.  Both are valid (neither is incorrect), but $O(b^d)$ is arguably more useful in the settings that typically arise in AI.

In the algorithms community and CS theory community, folks there tend to like to count the running time as a function of the number of vertices and edges in the graph.  Why?  In that context, worst-case running time is what makes most sense; also, in the problems that are typically considered in that community, in the worst case we need to examine the entire graph, so you typically can't hope to do better than $O(|V|+|E|)$.

In the AI community, folks tend to count the running time differently.  They often consider a specific kind of graph: a tree with branching factor $b$.  Also, in the situations that arise there, the graph is often infinite or very large.  Typically we try hard to avoid examining all of the graph -- that's often one of the major goals of the algorithms.  Thus, counting complexity in terms of $|V|$ and $|E|$ doesn't make sense: $|V|$ may be infinite, and in any case, we don't plan on examining all of the graph, so all that matters is the number of vertices we actually visit, not the number that may exist elsewhere but that we don't care about.

So, for the situations that often arise in the AI community, it's often more meaningful to measure the running time in terms of the branching factor of the tree ($b$) and the depth of the goal node ($d$).  Typically, once we find the goal node, the algorithm stops.    In such a tree, if we examine every vertex at depth $\le d$ before we find the goal node, we'll end up visiting $O(b^d)$ vertices before we stop.  Thus, if you like, you could think of this as visiting a subset of the graph with $|V|=O(b^d)$ (where now $V$ includes only the vertices we visit) and $|E|=O(b^d)$ ($E$ includes only the edges we look at), and you could think of an $O(b^d)$-time algorithm as one whose running time is $O(|V|+|E|)$... though this is a bit of an abuse of the $|V|,|E|$ notation.   Anyway, hopefully you can see why $O(b^d)$ is more informative than $O(|V|+|E|)$ in this context.