“Suppose p is prime. The theory of Abelian groups with all elements of order p has the extra axiom
(5) $px = 0$ “
I’m just trying to make sense of why this would even be the case. If we have an Abelian group with all elements of infinite order then how does infinite cardinality multiplied by a given x ever end up equaling 0...?
There are lots of abelian groups where every element has finite order. For example, any finite abelian group has this property, and there are lots of those. The trivial group - the unique (up to isomorphism) group with a single element - is the simplest example, but there are plenty of others. You're probably familiar with the integers modulo $n$ (for $n\in\mathbb{N}$, $n>1$) from basic number theory or combinatorics; this forms a group (called "$\mathbb{Z}/n\mathbb{Z}$") under addition modulo $n$, and every element of $\mathbb{Z}/n\mathbb{Z}$ has order dividing $n$. In particular, if $n$ is prime then every (non-identity) element has order $n$.
There are even infinite abelian groups where every element has finite order! Most simply, take any infinite direct power (= direct product where all the factors are the same) of $\mathbb{Z}/n\mathbb{Z}$ for some fixed $n$. When $n$ is prime, we get an infinite abelian group where every non-identity element has order $n$.
A bit more interestingly, any infinite direct sum of finite abelian groups has this property, although it may exhibit elements of wildly different finite orders. In particular, the group $$\bigoplus_{n\in\mathbb{N},n>1}\mathbb{Z}/n\mathbb{Z}$$ has every element of finite order and elements of every finite order.
I can't resist ending with my personal favorite example: the group $$\mathbb{Q}/\mathbb{Z}.$$ Intuitively, elements of this group are rational numbers in $[0,1)$, and addition is given by "wrapping around" just like in modular arithmetic (so e.g. ${1\over 2}+{2\over 3}={7\over 6}-1={1\over 6}$ in this group).
$\mathbb{Q}/\mathbb{Z}$ has a number of cute properties. It's an infinite abelian group where every element has finite order (think about $q\cdot {p\over q}$), but there are elements of every finite order (think about ${1\over n}$ for $n\in\mathbb{N}$). Additionally, all its finitely-generated subgroups are generated by a single element (think about how $\langle {1\over 2},{1\over 3}\rangle=\langle{1\over 6}\rangle$) but the whole group is not finitely generated, and in particular this illustrates the limits of the classification theorem for finitely generated abelian groups. And more subtly, while it makes perfect sense as a group there's no way to define a good notion of multiplication on it. It's a counterexample to many reasonable-sounding conjectures about abelian groups.