Question: My thinking:

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

Thanks

Answer: 
Trying to exclude $1+1=a$ and $1+1=b$ is very hard if you don't understand the following. All I am using here are field axioms, the fact that $0 \cdot x = 0$, and the notion of "order of an element in a group".

$(F , + )$ is a group of order 4, so there exists some element of order 2. Let $x \in F$ be such element. $x$ has order 2 means that $x \neq 0$ and
$$x+x=0$$
Hence
$$(1+1)= (x^{-1} \cdot x) (1+1) = x^{-1}(x \cdot (1+1)) = x^{-1} \cdot (x+x) = x^{-1} \cdot 0 = 0$$
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