Question: My thinking:

Due to the axiom describing closure of a field we know that there are four possibilities for the result.

Thanks

Answer: 
By Lagrange's theorem applied to the additive group of the field we have
$$
1+1+1+1=0.
$$
Therefore, by distributivity,
$$
0=(1+1+1+1)=(1+1)(1+1)=(1+1)^2.
$$
So if $1+1\neq0$ we have found a zero divisor $1+1$. Fields don't have zero divisors, so this is a contradiction and we are left with $1+1=0$.
The elements $0,1,a,b $ form an additive group. Either it is equivalent to $\mathbb Z\_4$ and it is equivalent to $\mathbb Z\_2\times \mathbb Z\_2$. If it is equivalent to $\mathbb Z\_2\times \mathbb Z\_2$ then $1+1=0$

So we can assume it is equivalent to $Z\_4$ and we can relable $0,1,a,b$ as $0,1,2,3$ and $1+1=2,2+1=3,$ etc.

But we can't make a field of this.

$2\*2=2 (1+1)=2+2=0$. But that would mean $1=\frac 12 \*2\*\frac 12\*2=\frac 12\*\frac 12 \*2\*2=\frac 12 \*\frac 12 \*0=0$. A contradiction.

.....

Oh, I guess I should point out that when we relable $0,1,a,b$ as $0,1,2,3$ we don't have to map $1$ to $1$. We could map $a$ or $b $ to $1$ and map $1$ to $2$ or $3$. But $2+2=0$ so if we map $1$ to $2$ we have $1+1=0$. If we map $1$ to $3$, we still have $3+3=2$ and a proof will still hold. 

If $0,1,a,b $ is a field and $1+1=x\ne 0$ then $x+x=0$ as that is the only additive group possibility. And $0=x+x=x (1+1)=x\*x $ which is impossible.

===

If $1+1=a $ then $1+b\ne 1,a,b$ so $1+b=0$ and $1+a=b $

$b+a\ne 0$ because $b+1=0$. $b+a\ne b $. $b+a\ne a $. So $ b+a=1$.

$a+a\ne a $, $a+a\ne 1$ because $a+b=1$, $a+a\ne b $ because $a+1=b $. So $a+a=0$

So $0=a+a=a (1+1)=a\*a $.

So $0\*\frac 1a =a\*a\*\frac 1a=a $

But $0\*x+0\*x=(0+0)x=0x $. So $0x=0x+0x-0x=0x-0x=0$.

$a=0\*\frac 1a =0$ which is a contradiction.

So $1+1\ne a $.

The exact same argument holds for $1+1\ne b$.