What do you call this property involving a function between two complete metric spaces? I have a notion, for which I am not able to find any reference name, as I am not that familiar with these concepts. Please help me by pointing to a definition for the below scenario.
Is there a name for the following property of the setup?
There is a a continuous and onto function $e : A \to B$, $A$ and $B$ being two different complete metric spaces.
For any element $b\in B$, and for any element $a \in e^{-1}(\{b\})$, 
(where $e^{-1}(\{b\})$ is the pre-image of the element $b$ in the function $e$),
For every punctured neighbourhood of $b$ denoted as $P_{\epsilon}(b)$, the pre-image $e^{-1}(P_{\epsilon}(b))$  contains a sequence $\{a_n\}$, such that $\{a_n\} \to a$  

Let us call the property described in question as Property P. Continuing the observations made in Stefan Böttner's answer we get the following.
Observation. Let $A$ and $B$ be metric spaces and $e\colon A\to B$ be a continuous function. Then $e$ has the Property P if and only if $A$ has no isolated points and $e$ is nowhere constant. (I am not sure to which extent this is a standard therm, but it seems to ba a natural name for this. It also appears in some books.)
By nowhere constant I mean that there is no non-empty open subset $U\subseteq A$ such that $e|_U$ is constant.
Proof. $\boxed{\Rightarrow}$ If $a$ is any point of $A$ then property P implies existence of a sequence converging to $a$, hence $a$ is not isolated.
Let $U\ne\emptyset$ and $a\in U$. Let $b=e(a)$. Let $\varepsilon>0$. The set $e^{-1}[P_\varepsilon(b)]$ contains sequence $(a_n)$ converging to $a$. Starting with some $n_0$, terms of these sequence belong to $U$ and we also have $e(a_n)\ne e(b)$. Therefore $e|_U$
$\boxed{\Leftarrow}$ Let $B(b,\varepsilon)$ be the open ball around $b$. By continuity we get that there is a $\delta$ such that $B(a,\delta)\subseteq e^{-1}[B(b,\varepsilon)]$. Let us choose $n_0$ with $1/n_0<\delta$. Then each ball $B(a,\frac1{n_0+k})$ lies inside $e^{-1}[B(b,\varepsilon)]$. And since the function $e$ is not constant on this ball, we can choose $a_k\in B(a,\frac1{n_0+k})$ such that $e(a_k)\ne e(a)$, i.e., $a_k\in e^{-1}[P_\varepsilon(b)]$.
