It is not the quotient of the direct sum of the rings. The product on the new ring is not defined in terms of the product on the direct sum.
Let's look at a sequence of rings $R_1\to R_2\to R_3\to \cdots$, with $\phi_{mn}:R_m\to R_n$ when $m\leq n$. I'm not sure if A-M uses the more general directed sets of rings, but the same argument can be made for that case.
Now, we construct a ring $R_{*}$ which has the additive group structure of $R_1\oplus R_2\oplus\cdots$, but the multiplicative structure is not the direct sum.
Intuitively, think of the elements of the group $(r_1,r_2,\dots)$ as representing the sum $r_1+r_2+\cdots r_n$. Then $r_i\times r_j$ is computed by moving $r_i$ and $r_j$ into the same ring - that is, if $i\leq j$ then $r_i\times r_j=\phi_{ij}(r_i)r_j$.
We then define the product in $R_*$ so that when $m\leq n$, $$(0,0,0,\dots,r_m,0,0,\dots)\times (0,0,\dots,s_n,0,0,\dots)$$
isthe element of $R_*$ which has all zeros except for the $n$th element, where it is $\phi_{mn}(r_m)s_n$.
This ring, $R_*$, has an identity, $(1,0,0,\dots)$.
Then you take $R_*$ modulo the ideal generated by the elements of the form $(0,0,\dots,r_n,-\phi_{n,n+1}(r_n),0,0,\dots)$.
Another approach would be to think of the additive group $R_*$ as elements of the form: $$r_0x^0+r_1x^1+\cdots r_dx^d$$ with $r_i\in R_{i+1}$.
Then we define the product so that $r_ix^i\times s_jx^j = \phi_{i,i+j}(r_i)\phi_{j,i+j}(s_j)x^{i+j}$. This yields a different product on $R_*$, but it still gives a ring.
In the general directed set case, $R_*$ won't even necessarily be a ring, because the product defined might not be associative, and $R_*$ might not have an identity, because the directed set in general does not have an initial element.
(If the directed set has an associative "join" operator, then the multiplication on $R_*$ can be made associative.)
Still the quotient will be a ring with identity. It's just much messier.
Best Answer
No. For instance, consider the ring $R=\mathbb{Z}[S]$ where $S$ is an arbitrary infinite set of indeterminates. Then any Noetherian subring $R_0\subset R$ can only contain finitely many elements of $S$ (otherwise the ideal generated by $R_0\cap S$ would not be finitely generated). It follows that $R$ cannot be a direct limit of a system of Noetherian rings indexed by $\mathcal{J}$ for any $\mathcal{J}$ of cardinality less than $|S|$, and so no fixed $\mathcal{J}$ can work for all $S$.