Characterization of fields in commutative rings. For the following, by "ring" I will refer to a ring with identity, and by "rng" to a ring wihout identity. By a rng-homomorphism I mean a ring homomorphism that does not necessarily preserves the identity (if it exists).

I know that if $R$ is a commutative ring, then the following are true:

*

*R is a field iff all non-zero rng-homomorphisms from it are injective.

*R is a field iff the only ideals of $R$ are $0$ and $R$.

My question is, is it necessary for $R$ to be both commutative and a ring? I.e., what if $R$
is just:

*

*A commutative rng.

*A (not necessarily commutative) ring.

*A (not necessarily commutative) rng.

Do $(1)$ and $(2$) still hold?
 A: Kind of reading between the lines here because the body of your question does not mention "fields" like the title does.  Anyhow, the two conditions are still equivalent even for (possibly noncommutative) rngs. Ideals correspond to kernels of departing homomorphisms, and the two points you listed are just two equivalent ways of saying that the only possibilities are the zero homomorphism or an injective homomorphism.
But of course they are not necessarily fields.
There are many examples of simple rings with identity that aren't fields, for example a square matrix ring over a field.
For an example of a noncommutative ring without identity that is simple, you could take the endomorphism ring $E$ of a countable dimensional vector space over a field, and then extract the subring $R$ of transformations with finite dimensional images.
Reasoning in that last example: the ring $E$ is known to have precisely three ideals, and you are selecting out the one nontrivial ideal.  Furthermore the ring is von Neumann regular, and it is known in such rings that "is an ideal" is a transitive relationship in the sense that if $I\lhd E$ and $J\lhd I$, then $J\lhd E$. So any ideal of $R$ would necessarily be an ideal of $E$ (and we know there are none between $R$ and $\{0\}$.)  It is not hard to produce two transformations that don't commute, and to see that the only candidate for the identity is the identity of $E$, which is absent owing to its infinite dimensional image.
There are also trivial examples of commutative simple rings without identity that (obviously) are not fields: say, the ideal $2\mathbb Z/4\mathbb Z$ of $\mathbb Z/4\mathbb Z$.
In short, the proposition you are looking at is interesting in the commutative case because it says "commutative simple rings are fields."  Simple rings are very diverse and interesting in general noncommutative algebra, but they are much more nicely behaved when commutativity is added.
