Let $p$ and $q$ be binary relation constants, such that $p(x, y)$ and $q(x, y)$ exist for all $x, y$.
We consider three objects: $a, b, c$. Let $p(a, b)$ and $p(b, c)$ be true, and all other $p$ be false.
Let us recursively define $q$ such that for all $x, y$: $$q(x, y) \Longleftrightarrow p(x, y) \vee \exists(t).(q(x, t) \wedge q(t, y))$$
Why is $q(c, a)$ consistent with the definitions of $p$ and $q$ above?
By the definition of $q$, $c$ must always be the first parameter to $p$, and $a$ must always be the second parameter to $p$. However, there is no such $p$ where this is true. $\neg\exists(y).(p(c, y)) \wedge \neg\exists(x).(p(x, a)).$
There are some oddities with your presentation. I don't know what you mean by "such that $p(x,y)$ and $q(x,y)$ exist for all $x,y$". Either $p$ and $q$ are fully defined on their domains, or they are not binary relations in the logic.
Your "definition" of $q$ is also not recursive. While you can think of the equivalence as modeling a recursion, it's just a constraint. This leads to why I used scare quotes around "definition". This does not uniquely define $q$. The minimal relation satisfying that constraint would be the transitive closure of $p$ and would contain only $(a,b)$, $(b,c)$, and $(a,c)$. From this, you would be right. But there is nothing stating that $q$ is the minimal relation satisfying that constraint. If $\forall x,y.q(x,y)$ held, your equivalence would also hold. The constraint only states that $q$ contains the transitive closure of $p$, not that it is the transitive closure of $p$.