Is there a specific name for this set of square-rooted primes? Consider the set of all the primes numbers (± square rooted) and all the irrational numbers that can be formed under their addition (only the addition of finitely many elements is allowed, i.e. no limits). We also include 0, as the identity element into this set. Is this set a group?
I believe yes. 
For example $7^{0.5}+3^{0.5}+5^{0.5}$ can be formed by adding $7^{0.5}+ 7^{0.5}+3^{0.5}+5^{0.5}+11^{0.5}$ and $-11^{0.5}$, (and infinitely many other other pairs of elements).
However, I have a doubt. Say we want to form $11^{0.5}$ from $5^{0.5}$ without resorting to the addition of $-5^{0.5}+11^{0.5}$. I believe that no finite sum of irrational primes (hence element) of the set exists such that $x=-5^{0.5}+11^{0.5}$. 
So now I wonder, what could this set be called (I haven't formally studied mathematics. Can it be regarded as a group, for I am not sure if it can fairly said to satisfy the latin square property. I am tempted to say that is does, but then again, is there a very specific name for the group I seem to have described, with its particular characteristic of "fundamental" elements and "constructed" elements?
Thank you very much. Please ask if any clarifications are needed.
 A: Your set is a group under addition and it is isomorphic to the additive group of maps $f\colon \mathbb N\to\mathbb Z$ where $f(n)=0$ for all but finitely many $n$; or to the direct sum of infinitely many copies of $\mathbb Z$.
But the most strikingly simple example of a group isomorphic to yours is:
$$\mathbb Q_{>0}, $$
the group of positive rational numbers (under multiplication).
The isomorphism is perhaps surprisingly straightforward: 
Given a natural number $n$ and a prime $p$ let $v_p(n)$ be the exponent of $p$ in $n$, i.e. we have $p^{v_p(n)}\mid n$, but $p^{v_p(n)+1}\nmid n$.
Then given any positive rational number $a=\frac nm$, map 
$$a\mapsto \sum_p (v_p(n)-v_p(m))\sqrt p.$$
Note that $v_p(n)=v_p(m)=0$ for all but finitely many primes, so that the sum is in fact finite. (And of course summands $k\sqrt p$ with  $k>1$ or $k<-1$ can be expressed as repeated sums of $\pm\sqrt p$)
Via this isomorphism, the $\sqrt p$ play the same "atomic" role for your group as the primes play for the multiplicative structure of $\mathbb Q_{>0}$: They are generators of the group, and these generators are free (no nontrivial relation holds among them) - which is a rephrasing of fact that your group / $\mathbb Q_{>0}$ os also isomorphic to the direct sum of infinitely many $\mathbb Z$'s, the free-abelian group on countably many elements, as mentioned above.
It should be mentioned that the following two facts make the isomorphim above valid:


*

*Unique prime factorization in $\mathbb Q_{>0}$: Any nontrivial product of prime powers is $\ne 1$

*Linear independence of prime roots: Any nontrivial rational linear coombination of square roots of primes is $\ne 0$


Prime factorization is Old Greek stuff, but the second factoid is modern and can be shown e.g. with  some Galois theory.
A: From your definition your set is closed under addition, it has a unique inverse for each element( for $a\in S, -a$ is the inverse) and it is also clearly associative. What your set does not have is an identity element, but this could be fixed by considering the union of your set and $\{0\}$.
