Every NP set is realized that way. And the characterization is correct in both directions if you add the stipulation that $f(b)$ and $b$ are polynomially related in size.

First, every NP set is a polytime image of a set in P. If a set $A$ is in NP, then there is a $P$-time decidable set $B$ of pairs (a,b) such that $a\in A$ if and only if there is $b$ such that $(a,b)\in B$. (And where the size of $b$ is polynomially restricted by the size of $a$ for all pairs in $B$.) Let $f$ be the function mapping (a,b) to a. Thus, $A$ is the image of $B$ under $f$, as desired.

~~Conversely, any P-time computable image of a P-time decidable set is in NP, since if $A$ is the image of $B$ under $f$, then $a\in A$ is witnessed by some $b$ such that $f(b)=a$.~~

Conversely, if $A$ is the image of an NP set $B$ by a function $f$ for which $f(b)$ and $b$ are polynomially related in size, then $a\in A$ is witnessed by $b$ such that $f(b)=a$.

Ryan's answer shows that an additional assumption on $f$ is required.