No: for example $F[x_1,x_2,\ldots]$ in countably many variables is commutative, has identity, and is not Noetherian.
There are holes in the reasoning you've given. The first one is
Now, $I=\cup_{i}I_i$ is a maximal ideal
This does not follow, even if the ring was Noetherian. In a Noetherian ring, that chain is always going to stabilize, and so the union will be the place where it stabilizes. But there's no reason to expect it will be a maximal ideal. The reason you gave: "Any ideal containing $\cup I_i$ falls in the increasing chain" does not hold. Consider the ring I gave above and the chain of ideals $(x_2)\subseteq (x_2,x_4)\subseteq (x_2,x_4,x_6)\subseteq \cdots$. The ideal $(x_1, x_2,x_3,x_4,\ldots)$ contains the union of the "even variable chain" but it is not one of the members of the chain, and it is not even equal to the union of the chain.
Secondly, even if the union is a maximal ideal, it does not help determine if the ring is Noetherian or not. One can construct non Noetherian valuation rings for which the union of all proper ideals is indeed the maximal ideal, but the ring is not Noetherian.
The condition is equivalent to the property that there is some cardinal $\kappa$ such that every left $R$-module is a direct sum of modules with at most $\kappa$ generators.
Theorem 2 of
Warfield, R. B. jun, Rings whose modules have nice decompositions, Math. Z. 125, 187-192 (1972). ZBL0218.13012.
shows that the commutative rings with this property are precisely the Artinian principal ideal rings.
For not necessarily commutative rings, Theorem 2.2 of
Griffith, P., On the decomposition of modules and generalized left uniserial rings, Math. Ann. 184, 300-308 (1970). ZBL0175.31703.
shows that all rings with this property are left Artinian.
There is a property of rings called (left) pure semisimplicity, which has several equivalent definitions. Section 4.5 of
Prest, Mike, Purity, spectra and localisation., Encyclopedia of Mathematics and its Applications 121. Cambridge: Cambridge University Press (ISBN 978-0-521-87308-6/hbk). xxviii, 769 p. (2009). ZBL1205.16002.
gives a good survey, including such results as:
(Theorem 4.5.4) A left pure semisimple ring is left Artinian, and every left $R$-module is a direct sum of indecomposable finite length indecomposable modules.
(Theorem 4.5.7) $R$ is left pure semisimple if and only if every left $R$-module is a direct sum of indecomposable modules, if and only if there is a cardinal $\kappa$ such that every left $R$-module is a direct sum of modules of cardinality less than $\kappa$.
(Note that one consequence of all of this is that if every left $R$-module is a direct sum of modules with at most $\kappa$ generators for some $\kappa$, then in fact every left $R$-module is a direct sum of finitely generated modules.)
So a complete answer to the question is that the rings in question are precisely the left pure semisimple rings, although one might argue that this is just replacing the original condition with an equally mysterious one.
But there is a conjecture (the Pure Semisimplicity Conjecture, which is 4.5.26 in Prest's book) that the left pure semisimple rings are precisely the rings of finite representation type (i.e., for which every module is a direct sum of indecomposable modules, and with only finitely many indecomposables). It is known that every ring of finite representation type is left pure semisimple. Also, "finite representation type" is a left/right symmetric property, so if the conjecture is true then pure semisimplicity is a left/right symmetric property.
Best Answer
The picture to have in mind for product rings (a picture which does not work for modules, by the way) is that a right ideal of $\prod_{i=1}^n R_i$ is always of the form $\prod_{i=1}^n T_i$ where each $T_i$ is a right ideal of $R_i$. This only applies to finite products. Containment works like you expect: $\prod_{i=1}^n T_i \subseteq \prod_{i=1}^n T'_i$ iff $T_i\subseteq T'_i$ for all $i$.
You can use this to solve your problem in a number of ways.
Given an ascending chain of right ideals $_1T\subseteq {_2T}\subseteq \ldots$ (I'm putting the subscripts on the left so they don't look like the subscripts of right ideals of the product ring above), you can find, for each index $i$, an index $i_n$ such that $_1T_i\subseteq {_2T_i}\subseteq \ldots$ has stabilized after $i_n$ steps. Since there are only finitely many $i$, you can just take the maximum $i_n$, and that means all the chains are simultaneously stabilized, and hence the original chain has stabilized.
Alternatively, if $N$ is an $R$ submodule of $M$ and you believe that $M$ is Noetherian iff $M/N$ and $N$ are Noetherian, then you can work inductively by saying that if $R=R_1\times R_2$ where $R_i$ are both Noetherian, you can use the first observation to conclude that the $R$ submodules of $R_1$ are exactly $T_1\times\{0\}$ for each right ideal $T_1\subseteq R_1$, and similarly for $R_2$, so that both of the $R$ modules $R/R_1\cong R_2$ and $R_1$ are Noetherian $R$ modules (since their submodules match those of the $R_i$, which are both Noetherian.)
Your idea of checking by finite generation also works once you note that $T=\prod_{i=1}^n T_i$ is finitely generated as an $R$ module iff each $T_i$ is finitely generated as an $R_i$ module.