We have an associative ring with unity $R$, and $S \subset R$ is multiplicative, and satisfies the following 2 Ore Conditions:
- for all $r\in R$ and $s \in S$, there exists $\lambda\in R, t \in S$ such that $\lambda s = tr$
- $\forall \phi, \psi\in R$, if $\exists s \in S$ such that $\phi s = \psi s$, then $\exists t \in S$ such that $t \phi = t \psi$
We view $S$ as a category with $\text{Hom}_S(s,t)=\{\lambda \in R \ | \ \lambda s = t\}$, and we let the functor $S \rightarrow \mathcal{Mod-}R$ send an object $s \in S$ to the free rank $1$ right $R$-module spanned by the basis vector denoted by $[s^{-1}]$, and an arrow $\lambda \in \text{Hom}_S(s_1,s_2)$ to the homomorphism acting on this basis vector as $[s_1^{-1}] \rightarrow [s_2^{-1}] \cdot \lambda$
We write $S^{-1}R$ for the colimit of this diagram. I have to show that it is formed by classes of formal fractions $s^{-1}r$ modulo the relation $s_1^{-1}r_1 \sim s_2^{-1}r_2$ (existence of $\lambda_1,\lambda_2 \in R$ such that $\lambda_1 s_1 = \lambda_2 s_2 \in S$ and $\lambda_1 r_1 = \lambda_2 r_2$), and also define a structure of associative ring with unit on $S^{-1}R$
Here are my observations: I think the Ore Conditions turn $S$ into a filtered category. Therefore our colimit $S^{-1}R$ is a filtered colimit. Therefore, we have the fact that $$S^{-1}R=\left( \coprod_{s \in S} R \right)/\sim$$ Where $\sim$ is precisely the relation we're looking for in the question, and therefore by this we can conclude the fact that $S^{-1}R$ is formed by classes of formal fractions $s^{-1}r$ modulo the relation $s_1^{-1}r_1 \sim s_2^{-1}r_2$ (existence of $\lambda_1,\lambda_2 \in R$ such that $\lambda_1 s_1 = \lambda_2 s_2 \in S$ and $\lambda_1 r_1 = \lambda_2 r_2$).
Now, how do I define a structure of associative ring with unit on $S^{-1}R$? And if anyone could attest to the validity of my arguments above, and if incorrect, help me out, I'll be obliged
Your thinking is correct, but your approach is confusing due to the simultaneous use of the first and second Ore condition. Another way to organize the construction you're trying to do is the following (I leave the proofs to you, a reference is Bo Stenström's book "Rings of Quotients, An Introduction to Methods of Ring Theory").
Proposition 1. For any multiplicative set $S$ satisfying the first Ore condition, the subset $M_{(S)}=\{(s,m):rs=0$ implies $rm=0\}/_\sim\subseteq S\times M/_\sim$, where the equivalence relation $\sim$ is given by $(s_1,m_1)\sim(s_2,m_2)$ if $\lambda_1m_1=\lambda_2m_2$ and $\lambda_1s_1=\lambda_2s_2$ for some $\lambda_1,\lambda_2\in R$, is the filtered colimit of $\mathrm{Hom}_{R-mod}(\left<s\right>,M)$ indexed by the left submodules $\left<s\right>\subseteq M$ reverse ordered by inclusion.
Corollary. $M_{(S)}$ has the structure of an abelian group with $(s_1,m_1)+(s_2,m_2)\sim(ts_2,\lambda m_1+tm_2)$ where $\lambda s_1=ts_2$ with $t\in S$.
Proposition 2. For each $(s_1,r)\in S\times R$ and each $s_2\in S$, each $t\in S$ such that $\lambda s_2=tr$ for $t\in S$ determines an abelian group homomorphism $\mathrm{Hom}_{R-mod}(\left<s_2\right>,M)\to\mathrm{Hom}_{R-mod}(\left<ts_1\right>,M)$ sending $\left<s_2\right>\xrightarrow{\mu}M$ to the pre-composition $\left<ts_1\right>\xrightarrow{\rho|_{\left<ts_1\right>}}\left<tr\right>=\left<\lambda s_2\right>\xrightarrow{\mu|_{\left<\lambda s_2\right>}}M$. In other words, we have an operation $(s_1,r)\cdot(s_2,m)\sim(ts_1,\lambda m)$ where $\lambda s_2=tr$. Furthermore, this operation is well-defined and bi-additive on equivalence classes so that the induced $R_{(S)}\times M_{(S)}\to M_{(S)}$ makes $R_{(S)}$ into a ring and $M_{(S)}$ into an $R_{(S)}$-module.
Proposition 3. There is a natural identification of $M_S=\{(s,m):rs=0$ implies $trm=0$ for some $t\in S\}/_\sim\cong (M/t(M))_{(S)}$ where $t(M)=\{m\in M: tm=0$ for some $t\in S\}$. Furthermore, when the second Ore condition holds, we have $M_S=(S\times M)/_\sim$, and that $M_S$ is the localization $S^{-1}M$.
Remark. The advantage to this method is that it generalizes. One can show that the first Ore condition holds for a multiplicative set $S$ if and only if the the set $\mathcal F=\{$left submodule $I\subseteq R:I\cap S\neq\emptyset$ is a Gabriel topology, i.e. it satisifes
$I\in\mathcal F$ implies $(I:j)=\{r\in R: jr\in I\}\in\mathcal F$;
$I\in\mathcal F$ and $(J:i)=\{r\in R: ri\in J\}\in\mathcal F$ for each $i\in I$ implies $J\in\mathcal F$.
Then the above constructions work to define $M_{(\mathcal F)}$ as the filtered colimit of $\mathrm{Hom}(I,M)$ indexed by $I\in\mathcal F$, and $M_{\mathcal F}=(M/t_{\mathcal F}(M))_{(\mathcal F)}$ where $t_{\mathcal F}(M)=\{m\in M:(0:m)\in\mathcal F\}$. The second Ore condition is then a strengthened version of the condition that the natural map $R\to R_{\mathcal F}$ be flat.