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 $1+1=0$.
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