Ideals and normal subgroups are a way in which to code what is "really" going on: we are moding out by congruences. I go into great detail in this post about why we can only define the obvious group structure on the quotient when we mod out by a normal subgroup.
As I say in the comments, I think it is not a good idea to think that you are moding out by an ideal because you are "really" moding out by a submodule of the $A$-module $A$: because the $A$-module structure on $A/I$ is not the ring structure on $A/I$. To get the ring structure via module actions you need to use the $A$-module structure on $A/I$ to induce an $A/I$-module structure on $A/I$... and that requires you to understand the ring quotient $A/I$, not the module quotient. You end up with a chicken-and-egg problem.
Given an algebraic structure, that is, a set $A$ together with a family $\Omega$ of operations on $A$, we are looking for equivalence relations $\sim$ on $A$ that "respect the operations".
Definition. Let $A$ be a set, and let $\Omega$ be a family of operations on $A$. We say that an equivalence relation $\sim$ on $A$ is a congruence if and only if for every operation $\mu\in\Omega$, if $(a_i)_{i\in\mathrm{ari}(\mu)}$ and $(b_i)_{i\in\mathrm{ari}(\mu)}$ are families of elements of $A$ such that $a_i\sim b_i$ for each $i$, then $\mu(a_i)\sim \mu(b_i)$.
In semigroups, which are sets with a unique binary operation $\cdot$, we ask that if $a_1\sim b_1$ and $a_2\sim b_2$, then $a_1\cdot a_2\sim b_1\cdot b_2$.
In monoids, which are sets with a binary operation $\cdot$ and a nullary operation $e$, we ask that if $a_1\sim b_1$ and $a_2\sim b_2$, then $a_1\cdot a_2\sim b_1\cdot b_2$, and that $e\sim e$ (that latter is "free" because we have an equivalence relation, but I include it to keep in in the general context).
In groups, we also have a unary operation ${}^{-1}$, so we ask in addition that if $a\sim b$, then $a^{-1}\sim b^{-1}$.
(It is a "happy accident" that it turns out we only need congruences in a group to be semigroup congruences to also be group congruences, so we really just need to check that it respects multiplication; this is the analog of how to check that a map between groups is a group homomorphism we only check that it is a semigroup homomorphism and we get the rest "for free".)
In rings, we ask that both binary operations be respected (and if you require your rings to have a unity, that the other nullary operation be respected as well).
Congruences are precisely the equivalence relations on $A$ for which $A/\sim$ "inherits" the algebraic structure of $A$ by operating on representatives. E.g., for a semigroup, if $\sim$ is a congruence on $S$, then $S/\sim$ is a semigroup with the operation $[a]\cdot[b]=[a\cdot b]$.
In the special case of groups, any congruence is completely determined by the equivalence class of $e$; this is essentially for the same reason that you can "detect" injectivity of a morphism by checking the kernel:
Theorem. Let $G$ be a group. Then a congruence $\sim$ on $G$ is completely determined by the equivalence class of $e$.
Proof. Let $\sim$ and and $\sim'$ be two congruences on $G$ such that $[e]=[e]'$. I claim that $a\sim b\iff a\sim'b$.
Indeed, $a\sim b$ and $b^{-1}\sim b^{-1}$, hence $ab^{-1}\sim bb^{-1}=e$. Thus, if $a\sim b$, then $ab^{-1}\in [e]$. Conversely, if $ab^{-1}\in[e]$, then $ab^{-1}\sim e$, and since $b\sim b$ wee conclude that $a\sim b$. Thus, $a\sim b\iff ab^{-1}\in [e]=[e]'\iff a\sim'b$ (by the same argument). $\Box$
I go into a lot more detail in the post mentioned above, showing that congruences are precisely the equivalence relations that are subgroups of $G\times G$, and that they correspond to normal subgroups in the natural way (the equivalence class of $[e]$).
If you do the same thing with rings, you likewise get that if $\sim$ is a congruence, then the class of $0$ is not just a subgroup/subring, but in fact a two-sided ideal. The condition of being an ideal is what "codes" the fact that the equivalence relation respects multiplication: $a\sim b$ and $c\sim d$ implies $ac\sim bd$.
The fact that we can "code" congruences in groups and rings with "special substructures" is unfortunately misleading (same is true for vector spaces and modules, except that there as in the case of abelian groups all substructures are "special"); once you work with more general structures like semigroups, you recognize that not every congruence can be so coded.
This is similar to the phenomenon that to describe a group or a ring by generators and relations you only need to list relations of the form something$=e$ (or $=0$ in rings), because any relation $a=b$ can be written as $ab^{-1}=e$. If you want to describe monoids, say, by generators and relations, you need to allow for more general types of relations, of the form something=something else, even if neither something nor something else are the additive identity.
