Is there a topological difference between an electric monopole and an magnetic monopole? When we introduce magnetic monopoles, we have duality, i.e. invariance under the exchange of electric and magnetic fields. 
Magnetic (Dirac) monopoles are usually discussed using topological arguments. The electromagnetic field is infinite at one point and thus we restrict our description to 
$$ \mathbb{R}^3 - \{0 \} \simeq S^2$$
The effect of a magnetic monopole is that it changes the topology such that we do no longer have the trivial bundle $S^2\times U(1)$, but instead the principal bundle $S^3$. Expressed differently, a magnetic monopole is described by the Hopf map $S^3 \to S^2$. 
Why don't we need this construction for "electric monopoles", i.e. an electric point charge like an electron? The electromagnetic field is also singular at the location of the electric monopole and thus I would suspect that the same line of arguments holds. In addition, doesn't duality tell us that there is "no" difference between an electric and a magnetic monopole?
I've never seen a discussion in topological terms of an electric point charge like an electron and thus I was wondering, why these are always only introduced for magnetic monopoles. 
 A: This is best understood in terms of differential forms, but vaguely, the difference is that $\mathbf E = -\nabla \varphi$ is a gradient, but $\mathbf B = \nabla\times \mathbf A$ is a curl.
If $A$ is the gauge potential, the field strength is the associated curvature $F = dA + A \wedge A = dA$ where for the second equality I restrict the discussion to electromagnetism, i.e., a $U(1)$ gauge group, so that $A \wedge A = 0$.
In $3+1$ dimensions, choosing an inertial frame with time coordinate $t$, the 2-form F can be expressed as $$F = E \wedge dt + B \tag 1$$
defining the electric part $E$ and the magnetic part $B$.  Notice that $E$ is a 1-form while $B$ is a 2-form.
(This corresponds to the usual expression of the components of the field tensor, $$F_{0i} = E_i \qquad \epsilon_{ijk} F_{jk} = B_i$$
however, this identification of $E$ and $B$ with vectors applies only in 3 spatial dimensions and (1) is the proper generalization to $n+1$ dimensions.)
Now, we see that $dF = 0$ corresponds to $$dE \wedge dt + dB = 0$$ 
and in the static case, this means that $$dE = 0 \quad \text{and} \quad dB = 0$$
separately.
These are the statements that the 1-form $E$ and the 2-form $B$ are closed. A global scalar (vector) potential exists if they are exact, i.e., $E = -d\phi$ for a a 0-form (scalar) $\phi$, and $B = d\tilde{A}$ for a 1-form $\tilde{A}$. Closed is always necessary for exact, but that closed is sufficient for exact is a topological property. A closed $n$-form is exact if the $n$:th de Rham cohomology $H^n_\text{dR}$ of the space vanishes. 
Now, for $\mathbb R^3 -\{0\}$, $H^1_\text{dR} = 0$ (this is equivalent to being simply connected), but $H^2_\text{dR} \neq 0$. Thus, there's a difference between the electric and magnetic cases.
TL;DR: $\mathbf B$ is a curl but $\mathbf E$ is a gradient, and those are topologically different, and this is obscured by not thinking in terms of differential forms.
A: Well, one difference is the gauge 4-potential $A_{\mu}$ in E&M. 


*

*On one hand, an electric monopole charge (i.e. a charge distribution in the form of a Dirac delta distribution) is consistent with a (possible singular) gauge 4-potential $A_{\mu}$ (e.g. a Coulomb potential) at the monopole position ${\bf r}$. There is no need to work with a punctured topology $\mathbb{R}^3\backslash \{{\bf r}\}$. 

*On the other hand, a magnetic Dirac monopole is incompatible (even in a distributional sense!) with a gauge 4-potential $A_{\mu}$ at the monopole position ${\bf r}$. It is mandatory to introduce non-trivial topology and/or Dirac strings.
It should perhaps be stressed that actual magnetic monopoles (which are so far experimentally unobserved) are thought to be 't Hooft-Polyakov monopoles, not Dirac monopoles, cf. my Phys.SE answer here.
A: The difference between the two arises because Maxwell's equations, while looking perfectly "equal", actually are not all of the same nature when we phrase electromagnetism in terms of a potential. If you think of $F$ as the dynamical variable, then
$$ \mathrm{d}F = 0 \quad \mathrm{d}{\star}F = 0$$
in vacuum look perfectly symmetric, and you might imagine adding electric and magnetic current 3-densities (these are the Hodge duals of the standard 1-vector current densitites) $j_\text{el},j_\text{mag}$ to obtain
$$ \mathrm{d}F = j_\text{mag}\quad \mathrm{d}{\star}F = j_\text{el}.$$
These would be the "Maxwell equations" of electromagnetism with both magnetic and electric charges. However, this theory has a "problem" - it is rather difficult to write it as a least action formulation. There is one, due to Zwanziger in "Local-Lagrangian Quantum Field Theory of Electric and Magnetic Charges", see also this answer of mine, but it is rather unwieldy and unnatural, and it has to artifically double the d.o.f. by introducing both an electric and a magnetic potential and imposing their field strengths being Hodge dual at the level of the equations of motion. My linked answer also remarks that there is a way to get magnetic monopoles that are not topological in the way you are thinking of here - this question seems to be about the singular Dirac monopoles rather than the non-singular 't Hooft-Polyakov monopoles.
Much more natural is to have the magnetic charges vanish, i.e. $\mathrm{d}F = 0$. Then, locally, by the Poincaré lemma there exists a 1-form potential $A$ with $\mathrm{d}A = F$, and there is the rather natural Yang-Mills Lagrangian with $A$ coupled to a current yielding $\mathrm{d}{\star}F = j_\text{el}$ when $A$ is considered as the dynamical variable. The crucial observation is that in this Lagrangian formulation, $\mathrm{d}F = 0$ is not an equation of motion. It is the Bianchi identity simply following from defining $F$ to be the derivative of the potential $A$, and it is therefore impossible to couple the theory of the electric potential $A$ to a magnetic current. As you already mention, introducing magnetic monopoles into this gauge theory requires "topological trickery", where we have to exclude the position of the monopole from the spacetime we are considering in order to rescue $\mathrm{d}F = 0$ and thus the description in terms of $A$, see also this answer of mine.
Now, you might say that since we introduced $A$ based on Maxwell's equations and these are perfectly symmetric, there is nothing fundamental about the magnetic charge that makes it the one that is supposed to be described in this topological way instead of the electric charge. We can switch $F$ and ${\star}F$, i.e. change which we see as the fundamental quantity and which as the Hodge dual, and define instead the "default" state of our gauge theory to be one in which electric charges are absent, so that we have a magnetic potential $B$ with $\mathrm{d}\mathrm{d}B = \mathrm{d}{\star}F = 0$. 
But since electric charges are so plentiful in our everyday world while magnetic ones are not, this is terribly inefficient.
A: 
Why don't we need this construction for "electric monopoles", i.e. an
  electric point charge like an electron?

It is not a matter of need, but it is relevant to wander why this construction has not been taken seriously before. And this relates to your final comment.

I've never seen a discussion in topological terms of an electric point
  charge like an electron and thus I was wondering, why these are always
  only introduced for magnetic monopoles.

There is, indeed, an article discussing in topological terms electric point charges. "Charge quantisation without magnetic monopoles: a topological approach to electromagnetism". 
Full disclosure: I am the author of this article.  
Somehow when describing the electromagnetic field tensor $F$ as a differential $2$-form in Minkowski spacetime, the "electric field" pairs with the time differential, e.g. one can find this in the article "Two-forms on four-manifolds and elliptic equations" by Donaldson, or in the article "On some recent interactions between Mathematics and Physics" by Bott, or even in the answer of Robin Ekman.
And this differential $2$-form is assumed to represent the curvature of some connexion. The usual $U(1)$ gauge description of the theory.
A priori, there is no physical reason for those choices, specially if one takes into consideration that the split of the electromagnetic tensor into a electric and magnetic part is arbitrary and nonphysical (it does depend on a particular choice of reference frame).
However, as it is shown in 1, the lack of evidence for magnetic poles (and the success of Maxwell's theory that do not include them), as well as the quantised nature of electric charges (a experimental fact observed by Millikan and Fletcher in 1909, and easily reproduced in introductory laboratory courses) support and indicate that is preferable to consider the hodge dual of $F$ as the curvature of some connexion. This is equivalent to pairing the "magnetic field" with the time differential when comparing it with the other cited articles.   
Although there is no need to treat electric poles via topology in the similar fashion magnetic poles are treated, it might be relevant and the "right" way to look at them, as it provides an explanation for electric charge quantisation without the need to invoke magnetic poles and quantum mechanics.  
A: From the book "Topology, Geometry and Gauge Fields——interactions" by Gregory L. Naber, section 2.2 Electromagnetic Fields, page 55, the first Chern class satisfying Maxwell theory is trivial. i.e. the electric charge does not encode the topology of spacetime. 
