I had this argument that looks correct but the result is, in my opinion, weird and potentially not true.
"Let $R$ be a ring and $P$ be its prime ideal. Take $\overline{a} \in R/P$ s.t. $a\notin P$. Consider the $R$-module homomorphism $R\to R \overline{a}, r\mapsto \overline{ra}$. This map is surjective. If $r$ is in the kernel then $ra\in P,$ hence $r\in P$. THerefore the kernel is $P$. By the first homomorphism theorem we have $R/P \cong R\overline{a}$."
The thing is, I just proved that all nonzero submodules generated by 1 element of $R/P$ are isomorphic to $R/P$ itself. One such kind of module is simple modules. But clearly $R/P$ is not necessarily simple.
Am I wrong somewhere?
You are not wrong, $P$ prime means $A = R/P$ is an integral domain. For any integral domain $A$, the principal ideals of $A$ are isomorphic to $A$ as $A$-modules.
As you've observed, this means that if $A$ has a submodule, i.e., if $A$ is not a field, then every principal submodule of $A$ has proper submodules. A simple submodule of $A$ would indeed be principal, so it is a contradiction if $A$ has a proper simple submodule. This means $A$ either is itself simple, i.e., $A$ is a field, or $A$ has no simple submodules. For an example of the latter consider the ring $\mathbb Z$. This is a PID so every submodule of $\mathbb Z$ is of the form $n\mathbb Z$ for some $n$, and $n\mathbb Z$ has $2n\mathbb Z$ as a proper submodule, so $\mathbb Z$ has no simple submodules. Note that there do exist simple $\mathbb Z$-modules (e.g., $\mathbb Z/2$), they just are not submodules of $\mathbb Z$.