Is an ideal also a normal subgroup?

Yes, as an ideal is closed under addition and subtraction, it is a subgroup. Since a ring is moreover an abelian group under addition, every subgroup is normal.

EDIT: To explicitly answer your questions:

1) Since a subring is also a subgroup and addition is abelian, something must first be a normal subgroup in order to be an ideal.

2) Yes, well-definedness under addition does technically require being a normal subgroup.

Recall the situation for a group $G$ and a subgroup $H$.

We know if $H$ is a normal subgroup then one gets a factor group $G/H$, if it is not normal then not.

But let us recall more precisely what is the issue there. If $H$ is not normal one can still define the set $G/H$ (and also $H\backslash G$) but does not anymore have an induced composition law on it. So, $G/H$ is not a group, yet not "for the usual reason" that some of the propoerties of the composition law is not fulfilled but since there is no reasonable law to begin with.

Note if you have classes $gH$ and $g'H$ and you want to define $(gH) \cdot (g'H) = gg'H$ then you need $ghg'h' \in gg'H$ for all $h,h' \in H$, which leads to the condition $hg' \in g'H$ and further to $g'^{-1}hg' \in H$, that is exactly the property for $H$ being a normal subgroup.

Now, let us turn to rings. You have a ring $R$ and a subring $H$ and you wonder when can one define a ring-structure on $R/H$ (where $R/H$ is defined as classes $\mod H$ with respect to addition). For the additive structure there is never a problem as $(H,+)$ is a subgroup of $(R,+)$ and by commutativity it is normal, and everything is fine.

However, what about multiplication. How to define $(a+H) \cdot (b+H)$. Well, 'of course' as $ab+H$. But for this to be well-defined you need that $(a+h)(b+h') \in ab+H$ for all $h,h' \in H$ which is the case if (and in general only if) $ah' \in H$ and $hb \in H$, that is you need $aH \subset H$ and $Hb \subset H$. And, note that you also need that additional condition for commutative rings! (Of course then one of the two suffices.)

So, if you want a multiplication on $R/H$ then you can only consider $H$ such that $aH \subset H$ and $Ha \subset H$ for all $a \in R$ (and also $HH \subset H$ but this is given by being a subring and/or as a special case of the just mentioned property), just like when you want a multiplication on $G/H$ then you can only consider $H$ such that $g^{-1}Hg \subset H$ for all $g \in G$.

This analogy is what is pointed out there.

To answer you questions more specifically, the properties you need for $R/H$ to have a natural ringstructure are:

1. $(H,+)$ is a [normal] subgroup of $(R,+)$.

2. $HH \subset H$.

3. $aH \subset H$ and $Ha \subset H$ for all $a \in R$.

The first two are summarized as saying $H$ is a subring (but in fact you could drop 2 altogether as it is a special case of 3).

However, the reason that normal subgroups are mentioned there at that point is not 1. but that 3. is a condition analogous to "normal" in the group context.

Every ideal is the kernel of a ring homomorphism. Since every ring homomorphism is an additive homomorphism, every ideal is the kernel of an additive homomorphism, and so a normal additive subgroup.