Let $f : A \to B$ be a ring homomorphism. In the left $A$-module structure on $B$ that's induced by $f$, what is $Ann(B)$?
I believe the $A$-module structure induced by $f$ is $a \cdot b := f(a)b$, and so
$$\operatorname{Ann}(B) = \{a \in A \mid a \cdot b = 0 ~ \forall b \in B\} = \{a \in A \mid f(a)b = 0 ~ \forall b \in B\}$$
If we know that $B$ is unital, then we could actually show $\operatorname{Ann}(B)= \ker f$, since if $a \in \operatorname{Ann}(B)$, then $f(a)b =0$ for all $b \in B$, so that $b=1$ implies $f(a)= 0$, whence $a \in \ker f$, and the other set inclusion is obvious. If not, I don't know what else one can show. Is that all there is to answering this question, or am I missing something?
Everything you have written is good.
Yes, having identity lets you conclude that $\operatorname{Ann}(B)=\ker(f)$.
You cannot say that in general though: without an identity, it could be, for example, that $B^2=\{0\}$, and then possibly $A=Ann(B)\supsetneq \ker f$.
For example, you could have the ideal $A=(x)+(x^3)$ in $\mathbb R[x]/(x^3)$ and $B=(x)+(x^2)$ in $\mathbb R[x]/(x^2)$, and the map $f:A\to B$ be the obvious one $r+(x^3)\mapsto r+(x^2)$. The kernel is $(x^2)+(x^3)$ but everything in $A$ will annihilate $B$.