A ring is an ordered triple, $(R,+,\times)$, where $R$ is a set, $+\colon R\times R\to R$ and $\times\colon R\times R\to R$ are binary operations (usually written in in-fix notation) such that:
- $+$ is associative.
- There exists $0\in R$ such that $0+a=a+0=a$ for all $a\in R$.
- For every $a\in R$ there exists $b\in R$ such that $a+b=b+a=0$.
- $+$ is commutative.
- $\times$ is associative.
- $\times$ distributes over $+$ on the left: for all $a,b,c\in R$, $a\times(b+c) = (a\times b)+(a\times c)$.
- $\times$ distributes over $+$ on the right: for all $a,b,c\in R$, $(b+c)\times a = (b\times a)+(c\times a)$.
1-4 tell us that $(R,+)$ is an abelian group. 5 tells us that $(R,\times)$ is a semigroup. 6 and 7 are the two distributive laws that you mention.
We also have the following items:
a. There exists $1\in R$ such that $1\times a = a\times 1 = a$ for all $a\in R$.
b. $1\neq 0$.
c. For every $a\in R$, $a\neq 0$, there exists $b\in R$ such that $a\times b = b\times a = 1$.
d. $\times$ is commutative.
A ring that satisfies (1)-(7)+(a) is said to be a "ring with unity." Clearly, every ring with unity is also a ring; it takes "more" to be a ring with unity than to be a ring.
A ring that satisfies (1)-(7)+(a,b,c) is said to be a division ring. Again, eveyr division ring is a ring, and it takes "more" to be a division ring than to be a ring. (5)+(a)+(b)+(c) tell us that $(R-\{0\},\times)$ is a group (note that we need to remove $0$ because (c) specifies nonzero, and we need (b) to ensure we are left with something).
A ring that satisfies (1)-(7)+(a,b,c,d) is a field. Again, every field is a ring.
We do indeed have that $(R,+)$ is an abelian group, that $(R-\{0\},\times)$ is an abelian group, and that these structures "mesh together" via (6) and (7). In a ring, we have that $(R,+)$ is an abelian group, that $(R,\times)$ is a semigroup (or better yet, a semigroup with $0$), and that the two structures "mesh well".
We have that every field is a division ring, but there are division rings that are not fields (e.g., the quaternions); every division ring is a ring with unity, but there are rings with unity that are not division rings (e.g., the integers if you want commutativity, the $n\times n$ matrices with coefficients in, say, $\mathbb{R}$, $n\gt 1$, if you want noncommutativity); every ring with unity is a ring, but there are rings that are not rings with unity (strictly upper triangular $3\times 3$ matrices with coefficients in $\mathbb{R}$, for instance). So
$$\text{Fields}\subsetneq \text{Division rings}\subsetneq \text{Rings with unity} \subsetneq \text{Rings}$$
and
$$\text{Fields}\subsetneq \text{Commutative rings with unity}\subsetneq \text{Commutative rings}\subsetneq \text{Rings}.$$
As explained in the linked answers, the notion of a "field with one element" is a catch all term for a system of linked ideas and phenomena throughout algebra, which aren't (yet) described satisfactorily within our current axiomatic system.
So the field with one element is not in any sense a field, or even a set with extra structure, and any proposed definition solely along these lines misses the point, which is to find an algebraic framework that explains the (observed) phenomena of interest. For instance, weakening the field axioms may allow one to build a field with one element, but it doesn't solve the problem of explaining whats actually going on. See this question Why isn't the zero ring the field with one element?
As an example, the Weil conjectures for curves state that for a smooth algebraic curve $C$ over a finite field $\mathbb{F}_q$, the number of points of $C$ defined over $\mathbb{F}_{q^n}$ differs from $q+1$ by at most $2g\sqrt{q^n}$. This is an arithmetic problem, counting the number of solutions to equations defined over finite fields, but it has a beautiful solution using algebraic geometry, utilising the $2$ dimensional geometric object $C\times C$ to do intersection theory.
To my knowledge, one of the main driving forces behind the desire for a theory of a field with one element is to replicate this argument, viewing $\mathbb{Z}$ as an algebra over $\mathbb{F_1}$ (whatever this means), if there was a sufficiently developed theory that worked as expected, so we had an "intersection theory", then an analogous argument could be used to prove the Riemann Hypothesis, which is the analogue of the Weil conjecture for curves.
So a good theory of $\mathbb{F}_1$ would allow one to make sense of $\mathbb{Z}\otimes_{\mathbb{F}_1}\mathbb{Z}$, and would be sufficiently precise to develop intersection theory and the estimates needed to make the above argument work.
This is to say, the motivations are there, but finding the right definitions to encapsulate the properties we are after is absolutely the hard part.
Best Answer
Stimulated by the comments, I now realize that you’ve already seen many examples of fields. If you’ve had two years of high-school algebra, you know the fields $\Bbb Q$ of all rational numbers and $\Bbb R$ of all real numbers. But you know more.
It’s universal here in the States, at least, to ask students to “rationalize the denominator” in fractions like $$ \frac{1+3i}{2+4i}=\frac{(1+3i)(2-4i)}{(2+4i)(2-4i)} =\frac{14+2i}{20}=\frac{7+i}{10}=\frac7{10}+\frac1{10}i\,, $$ although teachers may not ask students to perform the last step. You see, though, that the complex numbers $\Bbb C$ of all $a+bi$ with $a,b\in\Bbb R$ are a field, and this example certainly lets you believe that the Gaussian numbers, all $a+bi$ with $a,b\in\Bbb Q$, also form a field.
I hope that in high school you also simplified fractions like $$ \frac{1+3\sqrt2}{2+4\sqrt2} =\frac{(1+3\sqrt2)(2-4\sqrt2)}{(2+4\sqrt2)(2-4\sqrt2)} =\frac{-22+2\sqrt2}{-28} =\frac{11-\sqrt2}{14}=\frac{11}{14}-\frac1{14}\sqrt2\,, $$ so that you now realize that the numbers of form $a+b\sqrt2$ (where now you must restrict $a$ and $b$ to rational numbers), also play nicely together to make up a field.
It would be wrong to say that fields are everywhere in Abstract Algebra, but if you know where to look, you’ll find them surprisingly often.