Is the direct product of the one-holed torus with a segment homeomorphic to the handlebody of genus 2?
Since they have isomorphic fundamental groups and are aspherical, by the Whitehead theorem they are homotopically equivalent. Moreover, it is easy to visualize the homotopy by retracting the both things to the wedge of two circles.
They have homeomorphic boundaries, and, I believe, they should be homeomorphic themselves, but I am a bit slow to visualize it. If they are, how to see the product structure in the handlebody?
To see the product structure in the genus 2 handlebody, start by viewing the genus 1 handlebody as $T = S^1 \times [0,1] \times [0,1]$. I like to visualize the first two factors $S^1 \times [0,1]$ embedded as an annulus in the polar coordinate plane, where $(\theta,t) \in S^1 \times [0,1]$ maps to the point $(r,\theta)$ with $r=t+1$.
Pick a closed interval $I \subset S^1$. On the boundary of $T$ consider the two discs $I \times \{0\} \times [0,1]$ and $I \times \{1\} \times [0,1]$. Form the quotient space where you identify those two discs, $(x,0,t) \sim (x,1,t)$, for all $i \in I$. The result is the genus 2 handlebody, clearly expressed with the desired product structure: the one-holed torus is what you get from the annulus $S^1 \times [0,1]$ by identifying $I \times \{0\}$ to $I \times\{1\}$ with $(x,0) \sim (x,1)$ for all $x \in I$; and then cross with $[0,1]$.