A very basic proof:
If $G$ has an element of order $5$ and an element of order $7$, say $a$ and $b$, then since $G$ is abelian, $ab$ has order $35$ and $G$ is cyclic.
So we want to rule out the possibility that $G$ has no elements of order $5$, or alternatively, no elements of order $7$, without recourse to Cauchy's Theorem.
Elements of order $5$ occur "$4$ at a time" (any element of order $5$ generates a subgroup of order $5$, and any two such subgroups intersect in just the identity, so that gives $4$ elements of order $5$ in each distinct subgroup of order $5$), and similarly, elements of order $7$ occur "$6$ at a time".
Now $G$ has $34$ non-identity elements, which is neither a multiple of $4$, nor $6$, so we cannot have elements of only one order among the non-identity elements.
P.S.: the condition that $G$ be abelian is unnecessary-any group of order $35$ is actually cyclic, but a proof of this will have to wait until you have access to more advanced theorems.