For infinite sets, $A$ having larger cardinality than $B$ already implies that $A$ is "much larger," and perhaps a good way to see this is to think of all the ways of making $B$ "larger" which don't increase its cardinality. E.g. taking the union with any set of equal or smaller cardinality, taking the Cartesian product with a set of equal or smaller cardinality, etc.
Perhaps you want to know if there are a lot of other cardinalities in between those of $B$ and $2^B$. This question can't be answered with the usual axioms of set theory. For more information, read about the generalized continuum hypothesis.
Edit: In your extended question, you mention that the usual argument only guarantees that the power set $2^A$ has "one more" element than the original set $A$: given a map $\phi : A \to 2^A$, there is at least one element not in the image of $\phi$. Call that element $x$. Certainly there is a bijection between $A$ and $A \cup \{x\}$, so repeating your argument shows there isn't a bijection between $A \cup \{x\}$ and $2^A$; at least one element is missed. So in fact $2^A$ has "two more" elements than $A$. Repeating this, one sees there are infinitely many elements missed. Moreover, by showing that $A \times A$ cannot be mapped onto $2^A$, there are at least "$|A|$ more" elements in $2^A$, and so on.
This is what I was trying to get at in my first paragraph. Of course, it has to be understood in the context that adding one element to an infinite set doesn't actually make it any larger in the sense of cardinality. To my mind, cardinality is a very crude measure of the "size" of an infinite set, and so any operation that enlarges it to an extent that it can be detected by cardinality must be a very dramatic enlargement indeed.