Edited to fix three-dimensional limit:
Knot groups are torsion free. (This follows from the fact that knot complements are $K(\pi,1)$'s. I don't know a more direct proof.) The groups you describe are the result of killing the center of the torus knot group. This gives the orbifold fundamental group of the base orbifold. So, for example
$\pi_1(S^3 - K_{p,q}) = \langle x, y \mid x^p = y^q \rangle$
and the center is generated by $x^p$. By the way, $x$ and $y$ have geometric meaning. They are the cores of the two solid torus components of $S^3 - T$. So the torus knot group is one of the early examples of a free product with amalgamation: $\newcommand{\ZZ}{\mathbb Z}$ $\ZZ \ast_\ZZ \ZZ$.
Ok, if we kill the center (corresponding to crushing the circle fibers of the given Seifert fibered space) then we get the group
$\langle x, y \mid x^p = y^q = 1 \rangle.$
This is the free product $\ZZ/p \ast \ZZ/q$ that you are interested in. As mentioned above, this is the orbifold fundamental group of the orbifold $S^2(p,q,\infty) = D^2(p,q)$: that is, a sphere with three orbifold points, of orders $p$, $q$ and infinity or equivalently a disk with two points of orders $p$ and $q$. Now you want to take a limit as $q$ goes to $\infty$.
There is a "limit" of knots: namely $\lim_{q \to \infty} K_{p,q} = L_{2,2p}$ the $(2,2p)$ torus link. We have that
$\pi_1(S^3 - L_{2,2p}) = \langle x, z \mid zx^p = x^pz \rangle.$
As expected, $x^p$ generates the center. Killing $x^p$ gives the group
$\langle x, z \mid x^p = 1 \rangle \cong \ZZ/p \ast \ZZ$
as desired. We can also take the base orbifolds of this sequence of Seifert fibered spaces. In the limit we get $S^2(p,\infty, \infty)$ or equivalently an annulus with one cone point of order $p$. Again, this has orbifold fundamental group $\ZZ/p \ast \ZZ$. A reference for much of this material is Peter Scott's beautiful paper on the eight Thurston geometries. Taking geometric limits of knots is typically done in the hyperbolic setting - an example in Thurston's book is that the $(2,q)$ twist knots limit to the Whitehead link as $q$ tends to infinity.
We can avoid orbifold fundamental groups by instead taking connect sums of lens spaces: for example, the connect sum of $\mathbb{RP}^3$ and $S^2 \times S^1$ has fundamental group $\ZZ/2 \ast \ZZ$. For that matter, we can get the same fundamental group by taking the one point union of $\mathbb{RP}^2$ and a circle, $S^1$. However the above answer is directly inspired by your question.