A necessary and sufficient condition for determining prime ideals in any ring? Definitions
Definition 1. A set $S$ equipped with two binary operations '$+$' and '$\cdot$' will be said to be a pseudo-ring if,

*

*$(S,+)$ is a monoid


*$(S,\cdot)$ is a semi-group


*$a(b+c)=ab+ac$ and $(b+c)a=bc+ba$ for all $a,b,c\in S$.
Definition 2. A pseudoring $(S,+,\cdot)$ will be said to be a pseudo-integral domain $$ab=0\implies (a=0)\lor (b=0)$$ where $0$ is the identity of the monoid.
Theorem
Let $R$ be a ring and $P$ be an ideal of $R$. Then $P$ is a prime ideal of $R$ iff $(S,+,\cdot)$ is a pseudo-integral domain where, $$S:=\{I+P:I\ \text{is an ideal of}\ R\}$$ and also wehere '$+$' and '$\cdot$' on $S$ are defined as, $$(I+P)+(J+P)=(I+J)+P$$$$(I+P)(J+P)=IJ+P$$
Proof

$\color{red}{\text{Observation 1.}}$ First we need to show that operations on $S$ are indeed well-defined. We will just discuss the well-definedness of '$+$'. The well-definedness of '$\cdot$' follows in almost similar manner.
So, to prove the well-definedness of '$+$' let $I_1+P=I_2+P$ and $J_1+P=J_2+P$ for some $I_1,I_2,J_1,J_2\in S$ (observe that $S\ne\emptyset$ since $P+P\in S$). We need to show that, $$(I_1+P)+(J_1+P)=(I_2+P)+(J_2+P)$$To show this observe that, \begin{align}(i+p_1)+(j+p_2)\in (I_1+P)+(J_1+P)&\implies i+p_1\in I_1+P\land j+p_2\in J_1+P\tag{1}\end{align} Since $I_1+P=I_2+P$ we have $i+p_1\in I_2+P$. Similarly, $j+p_1\in J_2+P$. From this we can easily prove that, $$(i+j)+(p_1+p_2)\in (I_2+J_2)+P\tag{2}$$From $(1)$ and $(2)$ we conclude that, $$(I_1+P)+(J_1+P)\subseteq(I_2+P)+(J_2+P)$$The reverse inclusion can be done in a similar manner. Thus we have shown that '$+$' is indeed well-defined on $S$, the well-definedness of '$\cdot$' can be proved similarly.
$\color{red}{\text{Observation 2.}}$ After showing that the '$+$' and '$\cdot$' of $(S,+,\cdot)$ are indeed well-defined, we observe that $(S,+,\cdot)$ is indeed a pseudo-ring .  (The details are straightforward enough and hence is skipped).

After making these two observations we now proceed to prove our theorem. By definition of a prime ideal $P$ in $R$ we know that, $$P\ \text{is prime ideal of R}\iff (IJ\subseteq P\implies I\subseteq P\lor J\subseteq P)$$ where $I,J$ are two ideals of $R$.
Suppose that $P$ is prime in $R$. Then we will show that $S$ is a pseudo-integral domain. By $\color{red}{\text{Observation 2}}$ we can conclude that $S$ is a pseudo-ring so the only thing remaining to show is that if $(I+P)(J+P)=P$ (observe that the additive identity of $S$ is $P$) then either $I+P=P$ or $J+P=P$. Note that, \begin{align}(I+P)(J+P)=P&\implies IJ+P=P\\&\implies IJ\subseteq P\\&\implies (I\subseteq P)\lor (J\subseteq P)\\&\implies (I+P=P)\lor (J+P=P)\end{align} Conversely, let $S$ be a pseudo-integral domain and let $IJ\subseteq P$ we need to show that $I\subseteq P$ or $J\subseteq P$.  Observe that,   \begin{align}IJ\subseteq P&\implies IJ+P=P\\&\implies (I+P)(J+P)=P\\&\implies (I+P=P)\lor (J+P=P)\\&\implies (I\subseteq P)\lor (J\subseteq P)\end{align}
Question
Is there anything wrong in my theorem or in the proof?
 A: I haven't checked every line (not a big fan of parsing long proofs with lots of symbolic notation) but the undertones ring true.
Before we begin: there is already established terminology for something that is like a ring whose underlying set is an abelian monoid rather than an abelian group: they are called semirings. Since addition of ideals in any ring (and any quotient ring of a ring for that matter) is abelian, you could adopt this terminology instead. The set of ideals of any ring forms a semiring under the usual addition and multiplication of ideals.
Using "pseudo-ring" is bad because it already has a different widespread meaning. 
There may also be a term for when the monoid is additionally not abelian, but a good subsitute if you don't find one would be "seminearring" or "nearsemiring," considering the definition of nearrings.
I would summarize your idea like this: 

$P$ is a prime ideal of $R$ iff the semiring of ideals of $R/P$ has no nonzero zero-divisors.

This follows simply from the following points:


*

*The ordinary definition of $P$ prime in $R$ ($xy\in P$ implies $x\in P$ or $y\in P$) is equivalent with the ideal-wise version ($IJ\subseteq P$ implies $I\subseteq P$ or $J\subseteq P$)

*The ideal-wise sum and product in any ring (in particular, $R/P$) results in a semiring structure. The equivalence in the point above asserts that this semiring has no nonzero zero divisors.
Perhaps breaking it up this way will help you check and refine your proof. Good luck.
