Commutative Artinian rings in general are finite direct products of local Artinian rings, and that has nothing to do with it being an algebra over a special ring.
Every idempotent of such a ring generates an ideal (which is actually a subring) $eRe$. Sometimes it's possible that $e$ splits into two smaller nonzero orthogonal idempotents: $e=f+g$ such that $fg=0$, whereupon $eRe=fRf\oplus gRg$ and the idea splits into two smaller ideals. Using the Artinian condition, you refine the idempotents until they cannot be broken down any more.
The result is a set of finitely many idempotents $e_i$ which cannot be written as a sum of two other orthogonal nontrivial idempotents, and $\sum e_i=1$. The resulting subrings (which are ideals) $e_iRe_i$ are local rings, and $\oplus e_iRe_i=R$.
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
Let $S\subset\mathbb{Q}[x]$ be the subring consisting of polynomials with integer constant term. Note that $(p)$ is a maximal ideal of $S$ with $S/(p)\cong\mathbb{F}_p$. Let $R$ be the localization $S_{(p)}$. It is clear that $R$ is a local domain of characteristic $0$ with maximal ideal generated by $p$ and $R/(p)\cong \mathbb{F}_p$ is finite.
I claim, however, that $R$ is not Noetherian. Indeed, consider the chain of ideals $(x)\subset(x/p)\subset(x/p^2)\subset\dots$. These inclusions are all strict since $1/p\not\in R$ so $x/p^{n+1}$ is not a multiple of $x/p^n$ for any $n$.
Here is a very general principle along these lines. Let $R$ be any non-Artinian ring. Then I claim $R$ has an elementary extension (over the first-order language of rings) which is not Noetherian. Indeed, add a sequence of constant symbols $x_0,x_1,x_2,\dots$ to the language and axioms saying $x_n$ is not in the ideal generated by $x_0,\dots,x_{n-1}$ for each $n$. These axioms are finitely satisfiable in $R$, since $R$ is not Artinian so there are arbitrarily long finite chains of ideals in $R$. So by compactness, there is an elementary extension of $R$ that satisfies all of them, and thus is not Noetherian.
In particular, applying this with (say) $R=\mathbb{Z}_{(p)}$ this gives an elementary extension of $\mathbb{Z}_{(p)}$ that is not Noetherian. The properties of being local, characteristic $0$, $p$ being in the maximal ideal, a domain, and $R/(p)$ having $p$ elements are all first-order, and thus are still true of this elementary extension.