4 my italics didn't work

I think the answer is yes, there is such a universal group. Let $G$ be the direct sum group $\bigoplus_{n \in \mathbb{N}} G_n$, where $G_n$ is $\mathbb{Z}$ if the $n$th Turing machine does not halt, and $G_n$ is cyclic otherwise. Just construct each $G_n$ by adding $0$, $1$, $-1$, ... until the $n$th Turing machine halts. Then make $G_n$ the smallest cyclic group it could possibly be. The key is that you can run everything in parallel and don't have to define addition (since I am thinking of an Abelian group) when you define the element (for example, $5$ may not be created yet when $2$ and $3$ are). You just have to eventually get around to it.

It is easy to see that you can enumerate the elements of this group, and enumerate the multiplication table. This enumeration can then be turned into the code for each element of the group.

Update: In my construction above, there is only one generator associated with each Turning machine, namely the group element with 1 in the $n$th coordinate, where $n$ is the index of the Turing machine. To have two generators (as was requested), the easiest modification would be to have $G_n$ look like the free group on two elements $a$ and $b$. If the associated Turing machine halts, then make $a$ and $b$ have finite order, leaving the rest free. Again, I think this is similar to Joel's new construction.

(I specialize more in computable analysis than algebra, but I imagine there are a number of standard groups representing the halting problem.)

I am not sure if there is a "natural" such group.

Update: I should also add that usually, when you have a property like that---i.e. you know when it holds, but don't always know when it doesn't hold---then you can make it code a universal Turing machine. Such sentences are called $\Sigma^0_1$ sentences.sentences.*

3 fixes grammer errors

I think the answer is yes, there is such a universal group. Let $G$ be the direct sum group $\bigoplus_{n \in \mathbb{N}} G_n$. Where , where $G_n$ is $\mathbb{Z}$ if the $n$th Turing machine does not halt, and $G_n$ is cyclic otherwise. Just construct each $G_n$ by adding $0$, $1$, $-1$, ... to it until the $n$th Turing machine halts. Then make $G_n$ the smallest cyclic group it could possibly be. The key is that you can run everything in parallel and don't have to define addition (since I am thinking of an Abelian group) when you define the element (for example, $5$ may not be created yet when $2$ and $3$ are). You just have to eventually get around to it.

It is easy to see that you can enumerate the elements of this group, and enumerate the multiplication table. This enumeration can then be turned into the code for each element of the group.

(I specialize more in computable analysis than algebra, but I imagine there are a number of standard groups representing the halting problem.)

I am not sure if there is a "natural" such group.

Update: I should also add that usually, when you have a property like that---i.e. you know when it holds, but don't always know when it doesn't hold---then you can make it code a universal Turing machine. Such sentences are called $\Sigma^0_1$ sentences.

I think the answer is yes, there is such a universal group. Let $G$ be the direct sum group $\bigoplus_{n \in \mathbb{N}} G_n$. Where $G_n$ is $\mathbb{Z}$ if the $n$th Turing machine does not halt, and $G_n$ is cyclic otherwise. Just construct each $G_n$ by adding $0$, $1$, $-1$, ... to it until the $n$th Turing machine halts. Then make $G_n$ the smallest cyclic group it could possibly be. The key is that you can run everything in parallel and don't have to define addition (since I am thinking of an Abelian group) when you define the element (for example, $5$ may not be created yet when $2$ and $3$ are). You just have to eventually get around to it.
Update: I should also add that usually, when you have a property like that---i.e. you know when it holds, but don't always know when it doesn't hold---then you can make it code a universal Turing machine. Such sentences are called $\Sigma^0_1$ sentences.