So, this could be wrong, as the question merely says to find what group the tensor product is isomorphic to, but I believe it must be trivial.
Regardless, doing so "without theorems" is a bit tricky, and I am somewhat unsure what I am meant to do here. My first thought is to try and use some form of bilinear property to say $\langle a,b\rangle=ab\langle 1,1\rangle =\langle b,a\rangle$, thus $\langle 0,p\rangle=\langle p,0\rangle=\langle 0,0\rangle$, and $\langle q,0\rangle =\langle 0,0\rangle $, so as $q$ and $p$ are distinct primes, for all $a\in\mathbb{Z}/p\mathbb{Z}$, there exists $a'=qa$, and similarly $b'=pb$ for $b\in\mathbb{Z}/q\mathbb{Z}$, so, with those pieces in mind: for any $a,b$
$\langle a,b\rangle=\langle qa',pb'\rangle=a'b'\langle q,p\rangle=a'b'\langle 0,0\rangle=\langle 0,0\rangle$, so the entire group is trivial.
Additionally, if instead $p=q$, I believe the tensor product would just return $\mathbb{Z}/p\mathbb{Z}$ or also 0, and I am leaning towards 0 as well in this case, as we may say $p\langle a,b\rangle=\langle 0,b\rangle =\langle a,0\rangle$, thus $\langle a,b\rangle =\langle a,0\rangle +\langle 0,b\rangle =\langle a,0\rangle +\langle a,0\rangle =\langle 2a,0\rangle =2\langle a,0\rangle =\langle a,0\rangle =a\langle 1,0\rangle =\langle 1,0\rangle =\langle 0,1\rangle $ thus any tensor is equal to either basis tensors, meaning the group is trivial.
Am I handling this correctly? Most of my experience tensoring over $\mathbb{Z}$ was with nice theorems that said almost everything goes to 0.
Since $p\neq q$ we have $q^{-1}\in\Bbb Z_p$. $$a\otimes b=qq^{-1}a\otimes b=q^{-1}a\otimes qb=q^{-1}a\otimes 0=0$$