As mentioned in the comments, the term 'ring' came from Hilbert (previous question). Bill's answer there is excellent.
Fields are a bit funnier. It started with Dedekind using the word "Zahlenkörper" (body of numbers). In a supplement that he wrote to Dirichlet's Vorlesungenueber Zahlentheorie, he used that word (looking for an image for that) instead of 'rationally known quantities.' It was Moore who coined 'field' - in 1893, he wrote on Galois fields for the Bulletin of the New York Mathematical Society. However, he was always careful to mention 'fields of order' something, as field had the additional meaning at the time of neighborhood (as we understand it today).
I believe he used 'body of numbers' because they stay together just as a group does. But the transition to 'field' is a harder one to understand. Perhaps he meant it as an extension of the neighborhood-field idea - these numbers stayed in a certain neighborhood of a set of numbers in the sense that they were only a couple of powers of s away (referring to his passage below). Or perhaps he meant in to be similar to a 'body', in the sense that they just stay together.
In any case, if there were a strong, direct metaphor for 'ring' or 'field,' it is no longer readily apparent. The origins of the terms are difficult to track and harder to understand completely, and different people used the same terms in different ways. It is even possible that little thought was given to the naming of these things - it is most often hard to judge the importance of something new. Reaching for the term 'body' or 'field' because it fits and doesn't already mean something else might be the real origin.
Here is part of his piece on Galois fields for the Bulletin of the NYMS
Later, Edward Huntington wrote about the development of the theory of fields (it's funny how today we think of fields and groups as being basic, though groups are far older)
Here is an excerpt from Huntington's piece:
Huntington:
with footnote - indicating the well-understood meaning of field today:
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
$$\text{Fields}\subsetneq \text{Commutative rings with unity}\subsetneq \text{Commutative rings}\subsetneq \text{Rings}.$$
Best Answer
Among finite fields with characteristic $\neq 2$, David's example is unique. Specifically:
Proof: Suppose $F$ a finite field of characteristic $\neq 2$. Then $-1\neq 1\in F$. However, $(-1)\cdot (-1) =1$. Thus, the field $(F- \ \{0\}, \cdot, \ast)$ is characteristic $2$, so $|F -\{0\}| = 2^n$ for some $n \geq 1$.
On the other hand, the multiplicative group of a finite field is cyclic (see, e.g. this MSE question). Since the fields whose additive structure is cyclic are exactly the prime fields, we deduce that $n = 1$.
Thus, in this case, $|F| = 3$. This already implies that $F = \mathbb{Z}_3$. It remains to compute $\ast$.
We know that $\ast$ must have the property that $2\ast 2 = 2$ since $\ast$ maps $F -\{0,1\}$ to itself, and we also know that $1\ast 1 = 1$ because it's the additive identity of the field ($F-\{0\}, \cdot, \ast).$
Next, we see that \begin{align*} 2 &= 2\ast 2\\ &= 2\ast(1\cdot 2)\\ &= (2\ast 1) \cdot (2\ast 2)\\ &= (2\ast 1) \cdot 2,\end{align*} so $2 = (2\ast 1) \cdot 2$, so $2\ast 1 = 1$. Thus, $2\ast 1 = \min \{1,2\}$.
Likewise, \begin{align*}2\ast 0 &= 2\ast(0\cdot 2)\\ &= (2\ast 0) \cdot (2\ast 2)\\&= (2\ast 0)\cdot 2.\end{align*} If $2\ast 0\neq 0$, we can divide by it to learn $2 = 1$, which is absurd, so $2\ast 0 = 0 $. That is, $2\ast 0 = \min\{0,2\}$.
Moreover, \begin{align*} 1\ast 0 &= (2\cdot 2)\ast 0\\ &= (2\ast 0) \cdot (2\ast 0)\\ &= 0.\end{align*}
Finally, \begin{align*} 0\ast 0 &= 0\ast (0\cdot 1)\\ &= (0\ast 0)\cdot (0\ast 1)\\ &= 0,\end{align*} so $0\ast 0 = \min\{0,0\}$.
Thus, $a\ast b = \min\{a,b\}$ in all cases. $\square$
What about lonza's examples?
(I am not claiming lonza's operation $\ast$ is unique. I have no idea.)
Proof: Suppose $F_{2^n}$ is a finite field with $2^n$ elements. If $n = 1$, (i.e., $|F| = 2$), then $F-\{0,1\}$ is the empty set, so $(F-\{0,1\},\ast)$ doesn't form a group since it doesn't have an identity. Thus, we may assume $n\geq 2$.
Now, $F-\{0\}$ is again the underlying space of a field, so $|F-\{0\}| = p^m$ for some prime $p$. Thus, $p^m + 1 = 2^n$, so $1 = 2^n - p^m$.
By Mihalescu's Theorem, there is no solution to $2^n - p^m = 1$ with $m > 1$, so $m = 1$. Thus, $2^n - 1 = p$ is a Mersenne prime. $\square$
Finally, what about infinite fields?
Proof: As Geoffrey notes in his answer, $F$ must have characteristic $0$, so it contains a copy of $\mathbb{Q}$. Further, since $-1\neq 1$ and $(-1)^2 = 1$, $F-\{0\}$ must be characteristic $2$. This means that $a\cdot a = 1$ for any $a\in F - \{0\}$. But this is false when $a=2\in \mathbb{Q}-\{0\}\subseteq F-\{0\}.$