The universal cover of the trefoil complement is also the universal cover of a disk with three holes in it, cross an interval. Both are obtained from tilings of $\mathbb{H}^2\times\mathbb{R}$ by prisms which are ideal triangles cross intervals. The only difference is the way the two groups act on this tiling. Each takes two adjacent prisms as its fundamental domain. So the two are very closely related.
You can prove that
$$\mathbb R^3\setminus\{(0,0,z)\mid z\in\mathbb R\}\simeq \mathbb R^2\setminus\{(0,0)\}\simeq S^1,$$
where $\simeq$ means the spaces are homotopically equivalent. Since equivalent spaces have isomorphic fundamental groups, you're done.
Addition: Proving $A:=\Bbb R^2\setminus\{(0,0)\}\simeq S^1$: First we can define $i:S^1\to A$ to be just the inclusion map. Then take $r:A\to S^1$ to be
$$r(x)=\frac x{|x|}.$$
We want to show that $i\circ r\simeq \operatorname{id}_A$ and $r\circ i\simeq \operatorname{id}_{S^1}$. But $r\circ i$ is actually already equal to the identity function, so only the first one is left.
So, define $H:A\times I\to A$ by
$$H(x,t)=(1-t)\frac x{|x|}+tx$$
(for a fixed $x$, this is a straight line between $x/|x|$ and $x$). Note that $H$ is well defined and continuous. It is also easy to see that $H_0=i\circ r$ and $H_1=\operatorname{id}_A$.
Showing that $\Bbb R^3\setminus\{(0,0,z)\mid z\in\Bbb R\}\simeq\Bbb R^2\setminus\{(0,0)\}$ can be done in a similar way. Use the maps $r'(x,y,z)=(x,y)$ and $i'(x,y)=(x,y,0)$.
Best Answer
One nice method is to think of the trefoil as sitting on the surface of a torus. Think of it as slightly thickened. Then subdivide your space into a slight thickening of the torus (minus the thickened knot) and a slight thickening of the complement. The fundamental groups of each piece are isomorphic to $\mathbb Z$, while the intersection deformation retracts to the torus minus the trefoil, which is an annulus, so also has fundamental group $\mathbb Z$. If you take the loop generating the fundamental group of the annulus and push it into the torus, it winds around three times. If you push it out it winds around twice. So Van Kampen gives the following presentation $\langle x,y\,|\, x^3=y^2\rangle$.