Update. Here is a more direct construction. (See edit history for previous version.)
There is such a universal computable group as you request. Let $F$ be the free group on infinitely many generators $\langle a_p\rangle_p$, indexed by the Turing machine programs $p$. Let $G$ be the quotient of this group by the $k^{th}$ power $a_p^k$, whenever the program $p$ halts (on trivial input) in exactly $k$ steps.
Let us represent the group $G$ by reduced words in the generators $a_p$ and their inverses, but in the case that we took the quotient by $a_p^k$, then in these words we use exponents on $a_p$ in the interval $(-k/2,k/2]$. First of all, we can computably recognize whether a word in the generators fits this description, simply by checking whether it is reduced and whether any of the exponents is too large. The point of this last issue is that we can tell if the exponent $a_p^r$ is too large by checking if program $p$ halts in $2r$ steps or not. Similarly, we can easily compute the inverse of a word from the word, and we can computably multiply words. Again, whenever we have a word with some new exponents on the generators, we need to check whether they reduce because of our quotient, and this is possible by running the relevant computation for sufficient number to steps to determine it.
Thus, we have a computable representation of the group $G$.
Finally, I claim that it is universal in the sense you requested. Given any Turing machine program $p$, let $x_p=a_p$ and let $y_p=a_q$ for some other program $q$ known not to halt. Thus, by design, the group generated by $x_p,y_p$ will be the free group on these generators if and only if $p$ does not halt.
Lastly, let me observe that my group is not finitely generated, and it may be interesting to have a finitely generated example, or even a finitely presented example.