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.)
As I found out in this presentation, where the above-mentioned issue is dealt with in the context of default correlations, the problem is not possible to be solved with the inputs given.
I.e., when 3 events have defined probabilities of happening, the pairwise correlations are not enough to construct the joint probability of all 3 events happening.
In the context of default correlations, complex problems of correlations of events are solved using other methods such as Copula functions.
Best Answer
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