Following your and Stefanos' comments and your edit, I'm revising my answer. I think the first step is to precisely define what your events mean.
It seems like your probabilities $P_i$ are not the probabilities of event $i$ occurring in the sample space including all events. In a two event system, rather than $P_1=P(E_1)$, I think $P_1$ ought to be defined as the probability of event 1 happening in a world where all the other events can't happen. I.e.,
$$ P_1 = (E_1|\neg E_2), \text{ and similarly } P_2 = (E_2|\neg E_1) $$
What you want is $P(E_1 \cup E_2) = P(E_1) + P(E_2)$ because $E_1$ and $E_2$ are disjoint. By the theorem of total probability you can write $$P(E_1) = P(E_1|E_2)P(E_2) + P(E_1|\neg E_2)P(\neg E_2) = 0 + P_1(1-P(E_2)) = P_1(1-P(E_2))$$ And similarly $$P(E_2) = P_2(1-P(E_1))$$
Plugging these into the formula for one event or the other happening you get
$$\begin{align}
P(E_1 \cup E_2)
&= P(E_1) + P(E_2)\\
&= P_1(1-P(E_2)) + P_2(1-P(E_1))
\end{align}$$
Unfortunately you can't solve this for $P(E_1) + P(E_2)$ and the situation doesn't get any better as you add more events, although it does generalize easily. Nevertheless, your first step ought to be to define the problem correctly. For example you had
$P(E_2) = (1-P_1)*P_2$
whereas I think the following is correct based on my reasoning above
$P(E_2) = (1-P(E_1))*P_2$
and it's only by being very clear in the problem definition (which sadly means notation) that you can avoid subtle problems.
I've left some general comments from my original answer below in case they help.
events are not independent, specifically if one occurs all the following ones cannot occur
For the second event to happen event 1 must not have happened. So the probability of the second event is:
$P(E_2) = (1-P_1)*P_2$
This is not correct. $P(A \text{ and } B)=P(A)P(B)$ if and only $A,B$ are independent.
Question 1 ... the combined probability of exactly one event occurring will be simply the sum of the probabilities of all the events?
yes, exactly right, for any disjoint events $P(A\text{ or }B) = P(A) + P(B)$, as in Nicholas R. Peterson's comment.
(Note the general rule regardless of whether $A$ and $B$ are disjoint is $P(A\text{ or }B) = P(A) + P(B) - P(A\text{ and }B)$ which is easiest to see on a Venn diagram.)
Much of the post is not on the right track. The part that is on the right track is the list of three conditions. You need to show that the collection of all subsets of a (finite) set satisfies these conditions. The verifications will be in each case very short.
For example, is $S$ in our collection? Sure, every subset of $S$ is in our collection, and $S$ is a subset of itself.
Best Answer
Say your sample space is $\Omega=[0,1]$, and you want want to define probabilities on the possible events $X$ according to a distribution function $F(x)$. You are thus looking for a function $\mathbb{P}$ that assigns a probability to every subset $X \subset \Omega$ such that
and such that the probabilities assigned by $\mathbb{P}$ are meaningful in the sense that they obey the usual laws for probabilities,
$\mathbb{P}(\emptyset)=0$, $\mathbb{P}(\Omega) = 1$, $0 \leq \mathbb{P}(X) \leq 1$ for all $X \subset \Omega$,
$\mathbb{P}(X_1 \cup X_2 \ldots) = \mathbb{P}(X_1) + \mathbb{P}(X_1) + \ldots$ for all disjoint sequences $X_1,X_2,\ldots \subset \Omega$ (disjoint means $X_i \cap X_j = \emptyset$ if $i \neq j$, and note that countably infinite sequences are allowed!).
Even for some very well-behaved $F$ (e.g. for $F(x)=x$, it's hard to imagine a more well-behaved function than this), this turns out to be not possible. There simply isn't a function $\mathbb{P}$ that assign each subset of $[0,1]$ a probability (i.e. a real number between 0 and 1) such that the requirements above are fulfilled.
But it is possible if we exclude certain very weird and hard to imagine sets. All of these sets require the axiom of choice to even construct them, so you may imagine them to be artifacts of mathematical set theory, and not sets that you ever want to actually compute a probability for (unless you're a set theorist, maybe). Such sets are called non measurable.
Vitali sets are examples of non measurable sets, but unfortunately their construction requires a bit of number theory.