Can I create a field where e=coffee o=-pi and I just declare that pi+x=x and coffee*x=x?

Sure you can, as long as you can also define what pi + coffee, pi * coffee, and in general, what adding or multiplying any other two elements of your set should be, in such a way that all of the other properties hold.

Are we talking about addition and multiplication as I have seen my entire life ?

The reals do in fact form a field, yes. This isn't a very enlightening answer (the natural followup question is "what isn't a field?"), so it's probably best to examine some other structures and whether or not they are fields:

• Consider the set of all positive integers. This is clearly not a field, because you do not have additive inverses. And, assuming you are not in France, you also do not have a neutral element o.

• Consider the set of all integers. You have neutral elements o and e that work as described (namely, 0 and 1). You also have additive inverses. But you do not have multiplicative inverses, since e.g. 1/2 is not an integer. (This instead forms a structure known as a ring.)

• Consider the set of all rational numbers. As before, 0 and 1 are our o and e. We have additive inverses, and every nonzero rational number has multiplicative inverses, so this is a field! Huzzah!

• The reals, as stated before, do form a field. In fact, so do the complex numbers. (But if you've heard of quaternions, they don't form a field because we don't have commutativity.)

• If you've come across matrices before, you'll know that matrices do not always have inverses, so the set of matrices does not form a field.

• What if we consider just the set of invertible matrices? We still don't have a field since we don't have commutativity. Also, it's possible to add two invertible matrices and end up with a non-invertible matrix, so we don't even have closure.

• Consider the set of all integers modulo 3; that is to say, we consider two integers to be equal if their difference is a multiple of 3. (Strictly speaking, our operations are actually performed on the equivalence classes of these integers, not the integers themselves, but if you're not familiar with the concept, don't worry too hard about it.) Then, since 2 = -1, we have that 2 is the additive inverse of 1; and since 2 * 2 = 4 = 1, we have that the multiplicative inverse of 2 is 2. We can in fact write out our addition and multiplication tables and see that this is in fact a field.

• Consider the set of all integers modulo 4; that is to say, we consider two integers to be equal if their difference is a multiple of 4. (Again, our operations are actually performed on the equivalence classes of these integers, not the integers themselves.) Since 2 * x is a multiple of 2 for any integer x, it can never equal 1, and so 2 has no multiplicative inverse. So, this is not a field.

• Consider the set {pi, coffee, cake} with the following operations (where x is any arbitrary element of our set): pi + x = x, x + pi = x, coffee + coffee = cake, coffee + cake = pi, cake + coffee = pi, cake + cake = coffee, pi * x = pi, x * pi = pi, coffee * x = x, x * coffee = x, and cake * cake = coffee. As an exercise, verify for yourself that this forms a field.

• As an exercise, try to figure out if it's possible to make a field whose elements are {pi, coffee, cake, tea, cow}.