Is Minkowski space usually a vector space or an affine space? When I visited Wikipedia's page on Minkowski space, it seemed to offer two definitions. The first defined Minkowski space a vector space. 
Then, in a later section, it says 

The section above defines Minkowski space as a vector space. There is an alternative definition of Minkowski space as an affine space which views Minkowski space as a homogenous space of the Poincaré group with the Lorentz group as the stabilizer.  See Erlangen program.

Are both definitions equally valid? Do physicists usually prefer to work with one interpretation versus another? Do they offer different benefits?
 A: If I understood you correctly, you want to know the difference between the two.
Take a look at this question (and answer): What are differences between affine space and vector space?
Short answer: the only difference is that affine spaces don't have a special $\vec{0}$ element. But there is always an isomorphism between an affine space with an origin and the corresponding vector space.
In this sense, Minkowski space is more of an affine space. But you still can think of it as a vector space with a special 'you' point.
A: The choice is exactly the same as for Euclidean space, which can be represented by the vector space $\mathbb R^3$, or by the associated affine space. Either works fine, and the only real difference is the existence, in the vector space, of a zero vector. In general, it is more formally correct to use the affine version, which has in-built formal support for translation invariance; however, this can be cumbersome and it is often cleaner to work with the vector-space version.
It's important to note that, even mathematically, affine and vector spaces are already pretty close to indistinguishable. More precisely, given an affine space $A$ with an associated vector space $V$ (i.e. such that a left action $l: V\times A\to A$ exists, with the appropriate identity, associativity and uniqueness properties),


*

*$V$ itself is also an affine space, since it acts on itself trivially, and

*$V$ can be recovered from $A$ by fixing some point $p$ to act as zero, so that each $v\in V$ is in one-to-one corresponding to the point $q\in A$ such that $q=p+v$.


It is therefore not that meaningful to "choose" between either definition, as the objects are so closely equivalent.
In practice, unless you explicitly want to deal with the translation invariance, you often take $\mathcal M$ to be a four-dimensional vector space. It is good physics practice, in any case, to distinguish between events and displacements, both of which can be elements of $\mathcal M$, but for which some actions - specifically, adding one event to another - don't make that much sense. If you do that distinction in practice - that is, if you follow common sense about what you should add to what - then you're essentially using the affine-space definition, without all the cumbersome formal machinery. It's simply a lot easier to remember that the origin, as an event, is not all that meaningful, and not go to the lengths of formally striking it off the books.
A: Physically speaking, there is no preferred origin in the spacetime of special relativity, therefore an affine space (equipped with a Lorentzian scalar product) is a better model than a vector space. The Lorentz group acts on the tangent space at each event, this space being isomorphic to the space of four-displacements. The whole invariance group is the Poincare' one which includes translations.
A: The confusion arises as in many cases no distinction is made between the manifold of space-time and its tangent bundle, or even a characteristic fibre of it. Flat space-time is assumed to be a pseudo-Riemannian manifold $M$ with a pseudo-metric tensor $\eta\in T^*M^{\otimes 2}$ defined everywhere. The Minkowski vector space is any fibre $T_pM$, which is usually described with orthogonal inertial frames. Such frames are linked by transformations of the group $O(1,3)$, i.e. the full Lorentz group. The metric structure on $M$ can be used to turn the manifold $M$ itself into an affine space, since lengths in this space are translation-invariant. Hence the full isometry group of the space-time manifold $M$ is not just the full Lorentz group $O(1,3)$ but the larger Poincaré group $O(1,3)\ltimes\mathbb R^4$, i.e. the semidirect product with the translation group $\mathbb R^4$.
