Does a Finite Field with Four Elements Actually Exist?

I think you are supposed to verify the axioms. The ones for addition and multiplication are easy-you can just point to the groups that they represent. As you say, distributivity is the hard one. Nominally, given that addition is comutative you have $24$ cases to check, four multiplies times six sums. Half of them are trivial because they are multiplying by $0$ or $1$. That only leaves $12$, which isn't so many. I would do them and declare victory.

A field with $4$ elements is easily built: consider the quotient field $\;\mathbf Z/2\mathbf Z[X]/(X^2+X+1)$. It is a field because the polynomial $X^2+X+1$ is irreducible over $\mathbf Z/2\mathbf Z$, as it is a quadratic polynomial with no root inAlso, as a $\mathbf Z/2\mathbf Z$-vector space, it has dimension $2$, hence its cardinality is $4$.

The tables you provide reflect the laws on this quotient. For instance $x(x+1)=x^2+x\equiv -1$ since $x^2+x+1\equiv0$, and $-1\equiv1$ in $\mathbf Z/2\mathbf Z$.

Simply observe that addition and multiplication as defined in the table is equivalent to addition and multiplication of integer polynomials in $x$ modulo $2$ and modulo $x^2+x+1$, and then observe that they commute in a certain sense. $ \def\zz{\mathbb{Z}} $

Namely, for any $f,g \in \zz[x]$ define:

  $f \oplus g = (f+g) \bmod 2 \bmod (x^2+x+1)$.

  $f \odot g = (f·g) \bmod 2 \bmod (x^2+x+1)$.

Then prove that for any $f,g,m \in \zz[x]$ we have:

  $f \bmod m \bmod m = f \bmod m$.

  $(f+g) \bmod m = ( f \bmod m + g \bmod m ) \bmod m$.

  $(f·g) \bmod m = ( ( f \bmod m ) · ( g \bmod m ) ) \bmod m$.

Together these can be easily used to show (for you to do!) that associativity and commutativity and distributivity of $\oplus,\odot$ on the set of polynomials $F = \{0,1,x,x+1\}$ is equivalent to associativity and commutativity and distributivity of $+,·$ on integer polynomials, which is very easy to understand intuitively and prove.