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.)
Yes, these events are independent. The definition of independent events only refers to the probabilities, not sample spaces: $P(A\cap B)=P(A)P(B)$.
You could argue that for any (nontrivial) events, conditioning on one event always changes the other's sample space. For example, if we roll two dice and define $A$ to be the event "the first die shows $6$" and $B$ to be the event "the second die shows $6$", then these are clearly independent. However, the real sample space for either event is all $36$ possible outcomes, and the event space for $B$ is $\{(1,6),(2,6),\ldots,(6,6)\}$. Once you condition on $A$, the sample space for $B$ changes to $\{(6,1),(6,2),\ldots,(6,6)\}$, and the event space changes to just $(6,6)$, but the probability hasn't changed and that is what makes them independent.
In a comment you suggested that the real sample space associated to event $B$ is just a set of size $6$ (the roll on the second die), and conditioning on $A$ doesn't change that. But you can do exactly the same thing with your initial example.
Divide the $80$ people into $20$ groups of $4$. Do this in such a way that the first $10$ groups all have four baseball players, one of whom also plays football, and the remaining $10$ groups have no baseball players but one football player each. The numbers you gave ensure this is always possible.
Now number the groups $1,\ldots,20$ and number the people in each group $1,2,3,4$, with the football player in each group being number $1$. You can choose a uniformly random person by choosing a number from $1,\ldots,20$ to pick a group, then independently choosing a number from $1,2,3,4$ to select a person from that group.
Now the event "plays baseball" is "the first number is at most $10$". The event "plays football" is "the second number is $1$". This is now exactly the same as the dice example - one event only refers to the first number and the other to the second.
Best Answer
Whenever you need to find the probability of at least one thing happening, you can instead ask "What is the probability that none of them happen?" and subtract from $1$ (since the complementary event to "none happen" is "at least one happens").
For each of the individual events, we find the probability it does not happen by subtracting the probability that it does happen from 1. We have
Since the events are independent, the probability no event happens is the product of the individual probabilities, which is $133/1000$.
Based on the calculation above $$ \Pr(\text{at least one event}) = 1 - \Pr(\text{none of the events}) = 1 - \frac{133}{1000} = \frac{867}{1000} = 86.7\%. $$