if $G$ only has one subgroup of order $p$ and one subgroup of order $q$, then $G$ is a cyclic group

Since conjugation by an element of $G$ is an automorphism of $G$, we have $\lvert gPg^{-1}\rvert=\lvert P\rvert$ and $\lvert gQg^{-1}\rvert=\lvert Q\rvert$ for all $g\in G$, so that $gPg^{-1}=P$ and $gQg^{-1}=Q$ for all $g\in G$ by uniqueness of orders. Hence $P,Q\unlhd G$.

It is well-known that $P\cap Q\le P$ and $P\cap Q\le Q$, so, by Lagrange's Theorem, the coprimality of $p$ and $q$ implies that $|P\cap Q|=1$; therefore, $P\cap Q=\{1\}$.

Since $p$ and $q$ are coprime, it is clear that $\lvert PQ\rvert=pq=|G|$; thus $PQ=G$ (as $PQ\subseteq G$).

To show $G$ is cyclic, observe that $P\cong\Bbb Z_p$ and $Q\cong\Bbb Z_q$. Then it is routine to show that

$$\begin{align} \Bbb Z_{pq}&\cong \Bbb Z_p\times \Bbb Z_q\\ &\cong P\times Q\\ &\cong G \end{align}$$

by the Chinese remainder theorem.

This can be proved without Sylow's theorem.

Let $H$ be subgroup of order $p$ and $K$ be subgroup of order $q$, with $p,q$ distinct primes.

Clearly $H\nsubseteq K$ and $K\nsubseteq H$. Hence $H\cup K$ is proper subset of $G$.

Take $x\in G\setminus (H\cup K)$. What can be order of $x$? It is a divisor of $pq$, and it is not $1$.

If $o(x)=p$, then $x$ should lie in $H$ (unique subgroup of order $p$), contradiction. Similarly $o(x)\neq q$.

Hence $o(x)=pq$, i.e. $G$ is cyclic.