I didn't know $(x)$ had other meanings than simple parentheses around a variable.
Parentheses are being used in these axioms purely to clarify the order in which operations are taking place. If you don't know an operation $\ast$ is associative, then something like $a\ast b\ast c \ast d$ makes no sense at all, since you only know what $a\ast b$ is for two things $a,b$. Parenthesis can be used to group pairs so that it's clear what to 'multiply' first: say $(a\ast b)\ast(c\ast d)$ or $a\ast(b\ast( c\ast d))$.
If you're running into something like $(a+0)$, then formally rewriting this as $(a)$ isn't a meaningful thing to do. The parenthesis are a statement about the operation that's taking place, and there's no operation within the string "$(a)$".
It would be just fine to say that $x+0=(x+0)$, in which case the parentheses are redundant and not saying anything.
Can't we only say that $(x+0)+0=(x+0)$ from the axioms?
Sure, you can: firstly the right hand side can be replaced by $x+0$ alone, so that you have
$$(x+0)+0=x+0$$
According to the axiom that says $y+0=y$ for all $y$, the above holds with $y=x+0=(x+0)$.
Never mind this stuff below this line
original answer back when we thought the OP was stuck on something harder
When defining the real numbers using equivalence classes of Cauchy sequences of rational numbers (as it appears you are doing, denoting the equivalence class of a sequence $s$ as $(s)$) every element of your model of $\mathbb R$ is an equivalence class of the form $(s)$ for some rational Cauchy sequence $s$.
These equivalence classes are equal or unequal as sets, and there is no ambiguity like "$s=(s)$" at this point. However, there comes the time to look for where the rationals went in this process. They aren't literally in the set of equivalence classes, but there ought to be something that looks just like them.
Then one gets the idea that if we have $q\in \mathbb Q$, the sequence which is constantly $q$, call it $\vec q$, is a good candidate for representing $q$. But this sequence is not an element of the set of equivalence classes either, so it is not itself our new representative. Our representative ought to be $(\vec q)$, which is an equivalence class of rational Cauchy sequences. One can check that the map $\mathbb Q\to \mathbb R$ taking $q\mapsto (\vec q)$ is an injective ring homomorphism, so that we have found a copy of $\mathbb Q$ inside our model of $\mathbb R$.
At this point we say "hey you know what, among friends let's just say $q=(\vec q)$ and make the notation simpler." Everyone agrees, and you now have a new convention that lets you identify the rationals in your model of the reals. The equality isn't a mathematical one, it's just a conventional one that identifies elements of $\mathbb Q$ with their images in your model of $\mathbb R$.
Once this construction is over, you don't even really need to refer to the representatives $(s)$ at all. It's mainly important for proving that the real number axioms hold for the model. After that, it is "your old friend $\mathbb R$ and you can just say "let $r\in \mathbb R$ rather than referring to this underlying construction.
Something very similar is often done when defining $\mathbb C$ as addition and multiplication on $\mathbb R\times \mathbb R$. The element $(0,1)$ gets relabeled as $i$, and it turns out that the elements of the form $(r,0)$ for $r\in \mathbb R$ look exactly like $\mathbb R$, so it's OK to confuse $r$ with $(r,0)$ afterwards.
And when defining $\mathbb Q$ as equivalence classes of relation on $\mathbb Z\times \mathbb Z\setminus\{0\}$, we confuse an integer $z\in \mathbb Z$ with the equivalence class of $\frac{z}{1}$, etc.
