Classification of groups whose any non-trivial element can be associate to another to generate all Question: What are the groups (possibly infinite)  $G$ satisfying the following property?  $$ \forall g \in G \setminus \{ e  \} \ \exists g' \in G \text{ such that } \langle  g,g' \rangle = G.$$
Examples: the cyclic groups, the groups $C_p \times C_p$, and more generally $C_p \rtimes C_q$ with $p,q$ primes.
 A: Many such groups have a similar structure that you mentioned as examples. (This is not a complete answer to your question.)

Claim. An Artinian group $G$ with non-trivial center satisfying
  $$ \forall g \in G \setminus \{e\},\ \exists g' \in G \text{ such that } \langle g, g' \rangle = G \tag{$\ast$} $$
  is a cyclic group or the semidirect product of cyclic groups.

Proof. Let us assume that $G$ is not cyclic and prove $G$ is the semidirect product of cyclic groups. Namely, we shall prove that there are elements $a, b \in G$ such that 
$$\langle a, b \rangle = G,\ \langle a \rangle \unlhd G,\ \text{ and } \langle a \rangle \cap \langle b \rangle = \{ e \}.$$ Take a random central element $a_0 \neq e$ of $G$. By $(\ast)$, there is an element $b_0$ of $G$ such that $\langle a_0, b_0 \rangle = G$. If $\langle a_0 \rangle \cap \langle b_0 \rangle = \{ e \}$ then there is nothing to do. If not, then take $e \neq a_1 \in \langle a_0 \rangle \cap \langle b_0 \rangle$. Again, by $(\ast)$, there is an element $b_1$ of $G$ such that $\langle a_1, b_1 \rangle = G$. Repeating this process, we have a descending series
$$ \langle a_0 \rangle \ge \langle a_1 \rangle \ge \langle a_2 \rangle \ge \cdots $$
if $\langle a_n \rangle \cap \langle b_n \rangle \neq \{ e \}$ for all $n \ge 0$. Since $G$ is artinian, there is some index $i \ge 0$ such that $\langle a_i \rangle = \langle a_{i+1} \rangle$. Then
$$ \langle a_i \rangle = \langle a_{i+1} \rangle \le \langle a_i \rangle \cap \langle b_i \rangle \le \langle a_i \rangle$$
and we have $\langle a_i \rangle \le \langle b_i \rangle $. However, as $G = \langle a_i, b_i \rangle \le \langle b_i \rangle$, we have $G$ is cyclic which condradicts to our assumption. Therefore, $\langle a_n \rangle \cap \langle b_n \rangle = \{ e \}$ for some $n \ge 0$.
