Yes, there do exist rings which aren't Noetherian but which do have ACC on primary ideals. An example is $\prod_{i\in\Bbb N} F_i$ where the $F_i$ are fields.
This is clearly not Noetherian, and because it is commutative and von Neumann regular, all of its primary ideals are maximal.
This is even more dramatic than the ACC really, since you cannot even have a chain of two primary ideals :)
1). Having identity does not factor into the proof. Using noncommutativity would require you to restate everything in terms of right or left ideals.
2) Is an ideal generated by an infinite set has in fact infinitely many generators, or does this just mean that such an ideal is simply generated by infinitely many elements, but not necessarily infinitely many generators? The second thing is more correct. Really you should probably be thinking in terms of "finitely generated" and "not finitely generated" (which is a little different from "infinitely generated".)
3) The proof of the on direction is not correct as we have been discussing in the comments. To get you started for the $\impliedby$ direction:
Assume all ideals are finitely generated and let $I_1\subseteq I_2\subseteq\ldots$ be an ascending chain. Let $I=\bigcup_{i=1}^\infty I_i$ be the union of these ideals (and prove it is an ideal, if you haven't already.
Now $I$ is finitely generated by $\{a_1, a_2,\ldots a_n\}$. The important step in logic takes place right here:
prove that there must be a $j\in \mathbb N$ such that $\{a_1, a_2,\ldots a_n\}\subseteq I_j$. Conclude that $I=I_j$, and that $I_k=I$ for all $k\geq j$.
The idea for $\implies$ is the right one, but it has a technical flaw: you do not establish that the chain is strictly increasing. You can do this inductively. (The other posts mentions a fault with assuming $I$ is countably generated; however, I only see that as a disagreement in notation, and what is written doesn't really say that. You can well-order whatever generating set and write the initial segment of countably many generators like that, IMO.)
Best Answer
A noetherian ring $R$ can also be characterized by the fact that every ideal of $R$ is finitely generated.
Suppose that there exists an ideal $I$ in $R$ which is not finitely generated. Let $x_1\in I$. The ideal generated by $x_1$, $(x_1)\neq I$. There exists $x_2\in I, x_2$ is not in $(x_1)=I_1$. Suppose defined $I_n$ such that $I_{n-1}\subset I_n\subset I$, $I_n$ is finitely generated. There exists $x_{n+1}\in I$, $x_{n+1}$ is not in $I_{n}$, write $I_{n+1}=I_n+(x_{n+1})$. The sequence $I_n$ does not stabilize, contradiction.