So in the theory of topological vector spaces, from the fact that the translation map is a homeomorphism (that is, bijective, continuous and inverse continuous) it follows that the topology is entirely determined by any neighbourhood system.
1.) I guess that follows because we can just translate any open set, but my question is, why does the translation map needs to be a homeomorphism for this to be true. Or, vice versa, what would happen if any of the three properties of a homeomorphism would not apply to the translation map?
2.) Rudin states in his book about functional analysis without proof, that a subset $E$ of the t.v.s. $X$ is open if and only if all translates $a+E$ are open. How can I see this is true?
If you get rid of any of those 3 properties, you're not working with a topological vector space anymore. But suppose you do, and you're in a vector space with a topology attached, but which doesn't make a topological vector space.
The inverse axiom can't be removed, as it's just a property of vector spaces. And because every translation has an inverse which is also a translation, you can either remove one or both of the others. If you remove both, the axioms no longer give you any relationsship between the vector space and the topology, so you just have a separate vector space and topological space which happen to be defined on the same set. If you remove neither, you're still in a topological vector space, so translations are still homeomorphisms. So there isn't really a sensible way of removing any of the three properties.
So it isn't really meaningful to ask if the properties need to be true for this to work. There isn't any alternative case where those properties don't hold which we can look at. Unless you have an idea for an alternative structure you're interested in.
As to your second question: this is just another way of saying that translations are homeomorphisms. $f$ is a homeomorphism is equivalent to '$E$ is open iff $f(E)$ is open, for any subset $E$'.