Exercise: a local von Neumann regular ring with identity is a division ring.
Proof: If $a$ is nonzero, $axa=a$ for some element $x\in R$, and $ax$ and $xa$ are nonzero idempotents. Since local rings only have trivial idemoptents, $ax=1=xa$. Thus every nonzero element is invertible.
For the purposes of the question, I suppose that "local" for a ring without identity means "has one proper ideal containing all proper ideals." Along with von Neumann regularity, this is enough to prove $R$ has an identity.
Suppose $R$ is VNR and $M$ is the unique maximal ideal. Select $a\in R\setminus M$. Then $a=axa$ and $ax=e$ is an idempotent. Clearly $ax\notin M$, for if it were, $axa=a\in M$.
Then $(e)$ is some ideal of $R$. Suppose $(e)\neq R$: then by our supposition of what "local" means above, $(e)\subseteq M$, contradicting $e\notin M$.
Therefore the only possibility is that $(e)=eR=R$, but then it is easy to see that $e$ is the multiplicative identity of $R$. At that point, the exercise above shows $R$ is a field.
It works with minor modifications in the noncommutative case. Now we suppose there is a proper right ideal containing all proper right ideals, and a proper left ideal containing all proper left ideals.
Use $e=ax$ and $e'=xa$ to argue the same way, and you wind up with $eR=R=Re'$ to get that $e$ is a left identity and $e'$ is a right identity, therefore $e=e'$ is the identity for the ring. The exercise above indicates $R$ is a division ring.
Best Answer
This answer is the same as Zev's, but perhaps stated more "conceptually". For what it's worth:
A commutative ring is von Neumann regular if for all $a \in R$, there is $x \in R$ such that $a^2 x = a$.
Here are two straightforward facts:
Fact 1: Every quotient of a von Neumann regular ring is von Neumann regular.
[The defining condition is an identity, and if an identity holds in a ring it holds in any quotient.]
Fact 2: An integral domain which is von Neumann regular is a field.
[Fact 2 is literally the first thing that springs to mind when I see the somewhat strange defining condition. What does $a^2 x = a$ mean? Well, if we're allowed to cancel the $a$'s, it means $ax = 1$!]
Thus if $\mathfrak{p}$ is a prime ideal in a von Neumann regular ring, $R/\mathfrak{p}$ is a von Neumann regular domain, hence a field, so $\mathfrak{p}$ is maximal.