The ring R is local if it has a unique maximal left ideal. Remember that the jacobson radical is the intersection of the maximal left ideals, and in this case J is our unique maximal left ideal.
Remember that J is automatically an ideal, so R/J is well defined. Let y be an element of R representing a nontrivial coset of J. In other words, y is nonzero in R/J. Let y and J generate the left ideal M. This is all of R, hence y has a left inverse in R/J. This holds for every y, hence R/J is a division ring.
Conversely, let J be the jacobson radical and assume R/J is a division ring. Suppose there is a maximal left ideal M beyond J, and let y lie in M-J. Since xy = 1 in R/J, y and J span all of R. This is a contradiction, hence J is already a maximal left ideal, and it is the only maximal left ideal.
By symmetry, R/J a division ring implies J is the only maximal right ideal. Thus a local ring R has a unique maximal left ideal, or a unique maximal right ideal, or a jacobson radical J such that R/J is a division ring. The definitions are equivalent.
Asserting a unique maximal ideal is no longer sufficient. Let R be the 2×2 matrices over the reals. This is a simple ring whose only proper ideal is 0, yet it has many maximal left ideals, one for each line in the plane passing through the origin. It is not a local ring.
Conversely, assume R minus its left or right invertible elements is an ideal J. There can be no left ideals above J, except for R, hence J is the only maximal left ideal and R is local.
If R minus its units is an ideal J, then J has no left invertible elements, else it would contain all of R, including the unit 1. The left invertibles are all units, R minus its left invertibles is an ideal, and R is local.
If the nonunits form an ideal then they certainly form a closed subgroup under addition. Conversely, let the nonunits form a subgroup J. If x is in J and xy is a unit then x is also a unit, which is a contradiction. Therefore xy remains a nonunit, and lies in J. This makes J an ideal, and we are back to a local ring.
Don't try the converse. The localization of Z about 2 is local, but the nonunits are not nilpotent. In fact the ring is an integral domain.
Let R be local and suppose x is not right invertible. Embed x*R in J, the jacobson radical. Since J is an ideal, x cannot be left invertible either. Therefore R is dedekind finite.
Don't try for the converse. A ring like Z is dedekind finite, with no nontrivial idempotents, yet it is not local.