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)
Consider {1,a,b} which is a group under multiplication. 1 is the identity under multiplication. ab is 1 under multiplication as argued in the above answer.

Now, consider {0,1,a,b} . Try to find the additive inverse of 1. 0 is not the additive inverse obviously. Suppose a or b is additive inverse. Say a is the inverse 

Then 1+a = 0
(Multiply with b )
1.b + a.b = 0.b
(Since, 1.b = b, 0.b = 0 and a.b = 1) 
b + 1 = 0
But additive inverse is unique. Hence a =b

But a$\neq$ b because they are distinct elements of the group .Contradiction

Similar contradictions if you start with b.So additive inverse of 1 is 1
Hence 1+1 = 0