How to explain/model P(A), for an UNFAIR die, with set theory? Here is how I understand probability of event A: event A is a set and some outcomes of the sample space are elements of said set. When we need probability of event A we just calculate fraction 
(number of outcomes of set A)/(Total number of all outcomes in given sample space). For example, if we roll a die and event A is "the number that the die shown is 6", then P(A)=1/6, because we have set {6} for event A and {1,2,3,4,5,6} for our sample space. 
But alas, my understanding isn't without problems. What if our die is unfair and there is, say, 2/7 probability of the die showing 6 after one roll? How do we represent it in this set model? At first I was tempted to say that we have set A {6,6} correspodning to event A and {1,2,3,4,5,6,6} corresponding to the sample space. But then I remembered that it's just CAN'T BE. Repetitions are allowed for multisets, but not for sets! But as far as I know, we don't use multisets when explaining probability through lenses of set theory. 
 A: I suspect that by "elementary events" you think of events that have the shape $\{\omega\}$ where $\omega$ is an element of the outcome space $\Omega$. 
In the situation where outcome space $\Omega$ is countable then usually all these sets are taken to be measurable and the probability measure will then be determined by the values that it takes on these events. For $A\subseteq\Omega$ we have: $$P(A)=\sum_{\omega\in A}P(\{\omega\})$$Note that this agrees with your remark in your last comment: "we use elementary events to define the probability of everything".
If $\Omega$ is finite then we may (or may not) be in the special case where $P(\{\omega\})=1/|\Omega|$ or equivalently all $P(\{\omega\})$ have equal probability.
(If $\Omega$ is countably infinite then this is not even possible)
In your last comment you asked: "how do we define probability of an elementary event?"
This is depending highly on the looks of the problem that we are modeling. E.g. by a fair coin we usually take $\Omega=\{1,2,\dots,6\}$ with $P(\{i\})=\frac16$ corresponding with the special case I mentioned.
In the situation that you sketched (probability on $6$ is not $\frac16$) then this model is not okay and the model mentioned in my comment works fine.
You call the fact that elementary events themselves have probability "troublesome", but it is not. If every singleton $\{\omega\}$ is an event then in the first place it is unescapable that they have probability (all events have). In the second place the freedom we have in defining the probabilities of these events makes it possible for us to build suitable models for real-life situations (like throwing a fair coin, or also throwing an unfair coin).

Caution: what I said about determination of $P$ by the probabilities on events like $\{\omega\}$ is not true in general. It is true though if $\Omega$ is countable and all its singleton subsets are taken to be measurable.
