Subdivide the universal covering space $\mathbb R^2$ in the usual manner as squares, with vertical lines $x=m$ and horizontal lines $y=n$ for integers $m,n \in \mathbb Z$.
To visualize the desired covering space, draw two vectors based at the origin: $v = \langle 3,0 \rangle$ corresponding to $a^3$; and $w = \langle 2,1 \rangle$ corresponding to $a^2 b$. Let $P$ be the parallelogram determined by the vectors $v,w$, which form two sides of $P$, the other two sides then being determined. Now glue opposite sides of $P$ to form a quotient space $S$. And as usual, gluing opposite sides of the unit square $Q = [0,1] \times [0,1]$ gives the covering space $T$.
You can then use the pattern of intersections of the parallelograph $P$ with unit squares $[m,m+1] \times [n,n+1]$ to define the desired covering map $Q \mapsto T$.
It is theoretically more straightforward to view this construction using orbit spaces of deck transformations. If I have $(a,b) \in \mathbb R^2$ let me use $\tau_{(a,b)}$ to represent the translation $\tau_{a,b}(x,y) (x+a,y+b)$. Thus we can think of $T$ as the quotient of $\mathbb R^2$ by the action of the deck group $\langle \tau_{(1,0)},\tau_{(0,1)} \rangle$ (with fundamental domain $[0,1] \times [0,1]$), so
$$T = \mathbb R^2 / \langle \tau_{(1,0)},\tau_{(0,1)} \rangle
$$
and we can think of $S$ as the quotient of $\mathbb R^2$ by the action of the deck group $\langle \tau_{(3,0)}, \tau_{(2,1)} \rangle$ (with fundamental domain $P$), so
$$S = \mathbb R^2 / \langle \tau_{(3,0)}, \tau_{(2,1)} \rangle
$$
This way, the desired quotient map $S \mapsto T$ can be precisely defined as the map
$$S = \mathbb R^2 / \langle \tau_{(3,0)}, \tau_{(2,1)} \rangle \mapsto = \mathbb R^2 / \langle \tau_{(1,0)},\tau_{(0,1)} \rangle = T
$$
that is induced by the identity map $\mathbb R^2 \mapsto \mathbb R^2$.
One subgroup of $\langle a, b\ |\ abab^{-1} = 1 \rangle $ that is isomorphic to $\mathbb{Z}\times \mathbb{Z}$ is the subgroup generated by $a$ and $b^2$. Indeed these elements commute: $ab^2 = ba^{-1}b = b^2 a$, and as this is a finite-index sub-group of a surface group it is also a surface group, so in particular it only has one relation. In general you could take $a^m$ and $b^{2n}$.
In an answer to a related question (Is there a non-trivial covering of the Klein bottle by the Klein bottle?) I gave a few families of subgroups and determined the total spaces of the corresponding cover space in those cases, maybe you will find it useful.
Best Answer
The $e_n : S^1 \to S^1, e_n(z) = z^n$, are covering spaces corresponding to the subgroups $n\mathbb Z$. Moreover, $e_\infty : \mathbb R \to S^1, e_\infty(t) = e^{it}$ corresponds to the trivial subgroup of $\mathbb Z$. It is the universal covering. Therefore
$e_2 \times e_3 : S^1 \times S^1 \to S^1 \times S^1$ corresponds to $2\mathbb Z \oplus 3\mathbb Z$.
$e_\infty \times e_\infty : \mathbb R \times \mathbb R \to S^1 \times S^1$ corresponds to the trivial group.