Take the product of rings $M = \Bbb{Z}/(2) \times \Bbb{Z}/(3) \times \dots$ over the primes or in general take any infinite set of quotient modules of a ring $R$ and form their product. It's true then that a copy of $R$ lies in the product. So $\Bbb{Z}$ lies in the infinite product, M.
Can any more be said about $M$? Can we identify all points in $M$ that don't lie in the isomorphic image of $\Bbb{Z}$, into a "point at infinity" and also introduce $\Bbb{Z} \cup \{\infty\}$, so that $\Bbb{Z} \cup \{\infty\} \approx M / \sim$ somehow?
$M$ is a little too big. The map $\mathbb{Z} \to M$ sends an integer to its residues $\bmod n$ for all $n$, and when $n_1 | n_2$ you're not requiring any compatibility between the residue $\bmod n_1$ and the residue $\bmod n_2$. A better version of $M$ requires this compatibility; what you get is the profinite completion of the integers $\widehat{\mathbb{Z}}$. A precise sense in which this is better is that it has a natural (compact Hausdorff) topology with respect to which $\mathbb{Z}$ is dense.
There are many (in fact, uncountably many) points in $\widehat{\mathbb{Z}}$ not lying in $\mathbb{Z}$ and it doesn't seem to me to be reasonable to identify them. In fact $\widehat{\mathbb{Z}}$ is the product over all $p$ of the $p$-adic integers.