Is it possible to produce a field with three operations?

Among finite fields with characteristic $\neq 2$, David's example is unique. Specifically:

Suppose $F$ is a finite field with char $F\neq 2$ which has an extra operation $\ast$ as described by the OP. Then $F = \mathbb{Z}_3$ with $a\ast b = \min\{a,b\}$.

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?

Suppose $F_{2^n}$ is a finite field of characteristic $2$ which supports an operation $\ast$ as described by the OP. Then $2^n - 1$ is a Mersenne prime.

(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?

No infinite field $F$ supports such an operation $\ast$.

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 $a=2\in \mathbb{Q}-\{0\}\subseteq F-\{0\}.$

I don't know whether this is possible in general, but we can find at least once case of a "generalized" field. Consider the field $(F=\mathbb{Z}_3,+,\cdot,*)$ where $+$ and $\cdot$ are defined as normal, and $a*b$ is defined as $\min\{a,b\}$, where $0<1<2$. This is clearly a commutative operation. Also, $F\setminus\{0,1\}$ is the trivial group over $*$, since we have $2*2=2$. Finally, it's a little tedious, but we can show that $*$ is indeed distributive over $\cdot$ as follows:

For $0*x$:

$$0*(a\cdot b)=0=(0*a)\cdot (0*b),$$

For $1*x$:

$$1*(0\cdot a)=1*0 = (1*0) \cdot (1*a),$$

$$1*(1\cdot a) = 1*a = (1*1) \cdot (1*a),$$

$$1*(2\cdot 2) = 1*1=1=(1*2)\cdot(1*2),$$

where we only need to consider $2\cdot2$ in the last since the other values of $a$ were covered in the previous cases.

For $2*x$:

$$2*(0\cdot a)= 2*0 = (2*0) \cdot (2*a),$$

$$2*(1 \cdot a) = 2*a =(2*1) \cdot (2*a),$$

$$2*(2\cdot2) = 2*1 = (2*2)\cdot(2*2).$$

Note that in this generalized field, $0$ is still absorptive; i.e. $0*a=a*0=0$ for any $a$.

If $\ p=2^n-1\ $ is a Mersenne prime, let $\ q\ $ be a primitive element of $\ \mathbb{F}_{2^n}^{^\times}\ $ and define: $$ (q^a)*(q^b)=q^{ab}\ . $$ for $\ a,b=0,1,\dots,p-1\ $. Then \begin{align} q^a*(q^bq^c)&=q^a*q^{b+c}\\ &=q^{a(b+c)}\\ &=q^{ab}q^{ac}\\ &=(q^a*q^b)(q^a*q^c)\ . \end{align} The field $\ (\mathbb{F}_{2^n}^{^\times},.,*)\ $ is of course just (isomorphic to) the field $\ \mathbb{F}_p\ $.

Uniqueness: In his answer, Jason DeVito raises the question of whether the operation * is unique in this case. Since $\ q\ $ could be chosen to be any of the $\ \varphi(p)=p-1\ $ primitive elements available, it isn't, of course, unique in the strictest sense, but it's not difficult to show that any such operation in such fields must be given by this definition for some primitive element.

Let $\ e\ $ be the $*$-identity. Then $\ e\ne0\ $ and $\ e\ne 1\ $ (the identity element for $\cdot\ $), therefore $\ e\ $ must be primitive, because the order of $\ \mathbb{F}_{2^n}^{^\times}\ $ is prime. Now, if $\ r,s\in\mathbb{F}_{2^n}^{^\times}\ $, then $\ r=e^a\ $ and $\ s=e^b\ $ for some $\ a,b=0,1,\dots,p-1\ $, and $\ r*s=e^a*e^b=$$(e^a*e)^b\ $ by the distributivity of $*$ across $\cdot\ $, and $\ e^a*e=e^a\ $ because $\ e\ $ is the identity for $*$. Therefore, we have $\ r*s=(e^a)^b=e^{ab}\ $