I've found that R in my question is commutative but how can I check if R is a field? 
Is R commutative? I answered yes because for both multiplication and addition R fulfills the commutative axiom. I didn't take the time to do every possible combination but I did write
$0+0=0=0+0$
$e+0=e=0+e$
and so on for addition and 
$(0)(0)=0=(0)(0)$ 
$(0)(e)=0=(e)(0)$ 
and so on for multiplication. By the way, is there any better way to explain what I answered for 'is R commutative'? I feel like I answered it thoroughly enough.
For the question is R a field I said no because according to the chart no multiplicative combination of e,b,c will equals 1. But I'm not too sure. I'd appreciate your help.
 A: Showing associativity and distributivity would take the most time, but the exercise says not to bother with those. To check commutativity it is enough to check that the multiplication table is symmetric across its diagonal - in practice, with small tables this can be done by eyeballing it.
Can you see why it's enough to check the multiplication table is symmetric across its diagonal?
To check if it's a field, you want to


*

*find the multiplicative identity; which of $e,a,b$ has as its corresponding row and column the exact same entries as the labeling row and column (on the top or on the far left, resp.)?

*check if the multiplicative identity appears in every row and column of the multiplication table (not counting the ones that are zeroed out).


Question: why are (1) and (2) enough to check $R$ is a field?
A: Remember that "$1$" is just special notation in a ring for the identity element. It could also be labeled with any symbol depending on the author. In this case, the author is deliberately not using $1$ so as not to give away which symbol is the identity. That is for you to uncover.
So, the symbol $1$ not appearing on the table is not a problem. Among the symbols $\{0,e,b,c\}$ there is only one which acts as an identity... see which one?
One way to resolve your question about being a field in the negative by finding an element which can't be inverted on the multiplication table (that is, you can't get the identity you found in the last step as a multiple.)
An alternative "trick" to see it isn't a field is to use this: in a field, $x^2=x$ implies $x$ is zero or the identity. Here, $0,e,b,c$ all satisfy $x^2=x$, so that is more than two! Try to prove this fact by supposing $x\neq 0$ and multiplying on the left by $x^{-1}$ on each side.
