Let $A$ be a ring and $P$ be a proper bilateral ideal of $A$. Show that $P$ is prime if, and only if, for every pair of elements $a, b \in A$ it is satisfied $aAb \subseteq P ⇔a \in P$ or $b \in P$.
Note. The concept of Prime Ideal that I am dealing with is: $I$ is said to be a prime ideal of $A$ if for any $I_1$ and $I_2$ bilateral ideals of $A$ it is held that: $I_1$$I_2$ $\subseteq$ $I$ $⇔$ $I_1$ ⊆ $I$, or, $I_2 ⊆ I.$ I only have the definition, can you help me with a path?
Note.Consider that $1 \in A$.
My attempt was: Given that $1 \in A$, $a1b \in aAb \subseteq I$. Thus, $ab \in I$. To prove that $a \in I$ or $b \in I$, I plan to assume without loss of generality that if $a \in I$ then $b \in I$, but I have no clear idea how to relate having $ab \in I$ and $a \notin I$.