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)
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
$$(1+1)= (x^{-1} \cdot x) (1+1) = x^{-1}(x \cdot (1+1)) = x^{-1} \cdot (x+x) = x^{-1} \cdot 0 = 0$$