Practically, I need to prove that $\Bbb{Z}_p$ is a field when $p$ is a prime number without using theorems of the ring theory or quotient fields results. In order to do that, the set $S$ is defined as:
$$S=\{0,1,2,3,...,p-2, p-1\} $$ For notation, let's define $[a]_p$ as the modulo operation where $a$ is the dividend and $p$ is the divisor. Let $x,y\in S$, then the following operations are defined: $$x\oplus y=[x+y]_p$$ $$x\circ y=[xy]_p$$
First let's notice that the addition is closed since $[x+y]_p$ always results in a positive integer that's less than $p$. Also, by the division theorem we know that there exist unique integers $q,r$ such that: $$x+y=pq+r$$ Where $r=x+y-pq$ is the remainder of the division $(x+y)/p$, which means $r=[x+y]_p$ is unique and exists. Similarly, it can be proved that $x\circ y$ is also closed, exists and it's unique. So operations over $S$ are well defined.
- Associativity:
Let $x,y,z\in S$, then: $$x\oplus (y\oplus z)=[x+[y+z]_p]_p$$ By the modulo distributive property, $[x+[y+z]_p]_p=[[x]_p+[[y+z]_p]_p]_p$, and by the modulo identity property: $[[x]_p+[[y+z]_p]_p]_p=[[x]_p+[y+z]_p]$. This means: $$x\oplus(y \oplus z)=[[x]_p+[y+z]_p]_p=[x+(y+z)]_p=[x+y+z]_p$$ On the other hand $$(x\oplus y)\oplus z=[[x+y]_p+z]_p=[[x+y]_p+[z]_p]_p=[x+y+z]_p$$ Which proves the associatity. Since this is becoming quite extense, I'll leave it there, but the product associatity can be proven in a similar way. I wanted to show this just to give an example of the types or properties that I'm allowed to use.
I'm having problems with showing that there are multiplicative inverses. I have found a result that says that if $(x,p)=1$ then the linear congruence $xm\equiv 1 \mod p$ has a solution. This would help me to prove that there exists multiplicative inverses for every element of $S$, however, I'm not sure how to prove that such solution $m$ exists within $S$ and it's unique.
Uniqueness is straight forward, by using what you have already proved, suppose $a \in \mathbb{Z}_p$ has two inverses $b, c$. Then,
$$ c = c\cdot1 = c(ab) = (ca)b = 1 \cdot b = b $$
As for the existence, by Bezout's identity, if $(m:n) = d$, there exists a (minimal) linear combination
$$ am + bn = d $$
in particular in your case, if $x\in\mathbb{Z}_p$, $(x:p) = 1$, and therefore exist $a,b \in \mathbb{Z}$ such that
$$ ax + bp = 1 $$
and looking at this equality modulo $p$,
$$ ax \equiv 1 (p) $$
which is equivalent to saying that $[a]_p$ is the (unique) inverse of $x$ in $\mathbb{Z}_p$.