An argument without Sylow and Cauchy goes as follows.
For $G$ a group of order $pq$, with $p,q$ primes (WLOG $p<q$), the center $Z(G)$ is trivial or equal to the whole $G$. In fact, suppose $|Z(G)|=p$; then there is some $x\in G\setminus Z(G)$, and hence $Z(G)<C_G(x)<G$; but then (Lagrange) $p\mid |C_G(x)|$ and $|C_G(x)|\mid pq$; namely, $|C_G(x)|=kp$ for some integer $k>1$, and $kp\mid pq$; so, $k\mid q$ and hence ($k>1$) $k=q$. Therefore $|C_G(x)|=pq$, namely $C_G(x)=G$: contradiction. Likewise if we assume $|Z(G)|=q$.
If $Z(G)$ is trivial, then every nontrivial element of $G$ has centralizer of order $p$ or $q$. Therefore, the Class Equation yields:
$$pq=1+k_pp+k_qq \tag 1$$
where $k_i$ is the number of conjugacy classes of size $i=p,q$. Since $\langle x\rangle=C_G(x)$ for every $x\in G\setminus\{e\}$, there are exactly $k_qq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$); but each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $k_qq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because $q\nmid p-1$).
Therefore, $(1)$ yields:
$$pq=1+k_pp+m'q(p-1) \tag 2$$
for some positive integer $m'$; but then $q\mid 1+k_pp$, namely $1+k_pp=nq$ for some positive integer $n$, which replaced in $(2)$ yields:
$$p=n+m'(p-1) \tag 3$$
In order for $m'$ to be a positive integer, necessarily $n=1$ (which in turn implies $m'=1$, though this is not relevant here). So, $1+k_pp=q$: but this is a contradiction if we assume $p\nmid q-1$. So, in that case, we are left with $Z(G)=G$, namely $G$ Abelian.
Now, $G$ has at most one subgroup of order $p$, because if they were two, say $H$ and $K$, then $HK$ would be a subgroup of $G$ of order $p^2$ (contradiction, as $p^2\nmid pq$). Likewise, $G$ has at most one subgroup of order $q$, because if they were two, say $H'$ and $K'$, then $H'K'$ would be a subgroup of $G$ of order $q^2$ (contradiction, as $q^2\nmid pq$). But then, there are at least $pq-1-(p-1)-(q-1)=(p-1)(q-1)\ge 2$ elements of order $pq$, and hence $G$ is cyclic.