Question: My thinking:

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

Thanks

Answer: 
Finite fields have finte non-zero characterisctic.

And characteristic of a field is a prime number.
Also characteristic of a field divides the order of the field.

Here order of the field is $4$, and the only prime that divides it is $2$, hence characteristic of this field is $2$, hence $1+1=0$ !!!

**Alternative way:**

Clearly, $ab=1$ as $ab$ can't equal $a$ or $b$(as then multiply by inverse to get a contradiction), also it can't equal $0$, as any field is also an integral domain and product of two non-zero elements is non-zero.

Since a field is also a group under addition, order of an element under addition divides the order of the field, now if order of the element $1$ is not $2$, then it's of course equal to $4$, hence $1$ is the generator of the field.

So $1+1+1+1=0$, W.L.O.G 

$a=1+1$, $b=1+1+1$

So $ab=1+1+1+1+1+1=0+1+1=1+1$

So $ab\neq1$ (a contradiction)
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 $1+1=0$.