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Consider the operation defined on $\mathbb{N}$ by $x*y=\gcd(x,y)$.

I have two subpoints to solve in this problem. The first is:

$i)$ Prove that $(\mathbb{N}, *)$ is a commutative monoid

After proving that $*$ is an operation on $\mathbb{N}$, proving commutativity, and proving associativity (this was a bit harder than the other things to prove), the only thing that I had to prove was that $(\mathbb{N}, *)$ has an identity element. Meaning I have to prove that there $\exists$ $e \in \mathbb{N}$ such that $x * e = e * x = x$, $\forall$ $x \in \mathbb{N}$. So I chose $e = 0$ as the identity element, since $x * 0 = 0 * x = x$, $\forall$ $x \in \mathbb{N}$.

Having these $4$ conditions, we can conclude that $(\mathbb{N}, *)$ is a commutative monoid. The second suboint is:

$ii)$ Show that $D_n = \{ x \in \mathbb{N}: \hspace{0.25cm} x\mid n \}$ is a stable subset of $(\mathbb{N}, *)$ and $(D_n, *)$ is a commutative monoid.

Through a bit of work, I managed to show the first part of this $ii)$. However, in the second part, I have to show that $(D_n, *)$ is a commutative monoid. We can prove that $(D_n, *)$ is an operation using the first part of $ii)$ (so $*$ is an operation on $D_n$ because $D_n$ is a stable subset of $(\mathbb{N}, *)$) and we can prove commutativity and associativity in a similar manner as we did in subpoint $i)$. The problem I have is with the identity element.

I used $0$ as the identity element in subpoint $i)$, but I can't use $0$ here, since $0$ is not a divisor of any number, so $0 \notin D_n$ for any $n \in \mathbb{N}$. So I can't use that as the identity element. I though of using $e = x$ as the identity element for every $x \in D_n$, since $\gcd(x, x) = x$. But then we wouldn't really have the condition:

$$\exists \hspace{0.2cm} e \in D_n \hspace{0.2cm} \text{such that} \hspace{0.2cm} x * e = e * x = x \hspace{0.2cm}, \forall x \in D_n$$

But, rather, we would have it the other way around:

$$\forall x \in D_n, \hspace{0.2cm} \text{there} \hspace{0.2cm} \exists \hspace{0.1cm} e\in D_n \hspace{0.2cm} \text{such that} \hspace{0.2cm} x * e = x * e = x$$

So I don't think this is right, since the condition met goes the other way around, although I am not sure.

So how could I prove that $(D_n, *)$ is a commutative monoid? $0$ is not among $n$'s divisors so I can't use $0$ as the identity element. But a commutative monoid needs an identity element. What am I missing here?

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Your doubts are justified; an element $e\in M$ is the identity element of the operation $\star$ if $$\forall m\in M:\quad e\star m=m\star e=m.$$ This is one element $e\in M$ that satisfies this relation for all $m\in M$. So indeed the fact that $$x\ast x=\gcd(x,x)=x,$$ does not tell you that $x$ is the identity element.

To find the identity element of $D_n$, the set of positive divisors of $n$, you need to find a positive divisor $e$ of $n$ such that $$\forall x\in D_n:\quad e\ast x=\gcd(e,x)=x.$$ In particular, you need $\gcd(e,n)=n$. That only leaves one option. Can you show that this is indeed the identity element of $D_n$?

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