In particular, for any infinite group $G$ at all, of any size, we may take a countable elementary subgroup $H$, meaning in particular that they have the same first-order theory, and so there is a nonprincipal ultrafilter $U$ on an index set $I$ such that the ultrapowers $G^I/U\cong H^I/U$ are isomorphic. Since every first-order structure maps elementarily into its ultrapowers, this means in particular that $G$ maps elementarily (and hence monomorphically) into an ultrapower of $H$, a countable group.
Theorem. For every group $G$ there is a countable group $H$ and a free ultrafilter $U$ on each a set, such that $G_i$ being countable).G$embeds into the ultrapower$H^I/U$. If you want to insist that the ultrafilter concentrate on index set$\mathbb{N}$, however, then things become more complicated. If the CH holds, then the Keisler-Shelah theorem shows that any two groups of size at most$2^{\aleph_0}$and with the same theory have isomorphic ultrapowers by an ultrafilter on$\aleph_0$, and so the desired result is attained. In the non-CH case, however, what we seem to get is that for any cardinal$\lambda$, if$\beta$is smallest such that$\lambda^\beta\gt\lambda$, then any two groups of size$\beta$with the same theory have isomorphic utrapowers using an ultrafilter on$\lambda$. Thus, they each map into an ultrapower of the other. The Keisler-Shelah theorem was proved first by Keisler in the case that GCH holds, using saturation ideas as in Simon's answer. The need for the GCH was later removed by Shelah. 2 Fixed Keisler spelling; added 9 characters in body The general situation, where CH fails, may be informed by the Kiesler-Shelah Keisler-Shelah isomorphism theorem, which asserts that two first-order structures have isomorphic ultrapowers if and only if they have the same first-order theory. In particular, for any infinite group$G$at all, of any size, we may take a countable elementary subgroup$H$, meaning in particular that they have the same first-order theory, and so there is a nonprincipal ultrafilter$U$on an index set$I$such that the ultrapowers$G^I/U\cong H^I/U$are isomorphic. Since every first-order structure maps elementarily into its ultrapowers, this means in particular that$G$maps elementarily (and hence monomorphically) into an ultrapower of$H$, a countable group. Thus, this fully answers the version of question 2 in which we allow the ultrafilter to live on a bigger index set (but still insist on each$G_i$being countable). If you want to insist that the ultrafilter concentrate on$\mathbb{N}$, then things become more complicated. If the CH holds, then the Kiesler-Shelah Keisler-Shelah theorem shows that any two groups of size at most$2^{\aleph_0}$and with the same theory have isomorphic ultrapowers by an ultrafilter on$\aleph_0$, and so the desired result is attained. In the non-CH case, however, what we seem to get is that for any cardinal$\lambda$, if$\beta$is smallest such that$\lambda^\beta\gt\lambda$, then any two groups of size$\beta$with the same theory have isomorphic utrapowers using an ultrafilter on$\lambda$. Thus, they each map into an ultrapower of the other. The Kiesler-Shelah Keisler-Shelah theorem was proved first by Kiesler Keisler in the case that GCH holds, using saturation ideas as in Simon's answer. The need for the GCH was later removed by Shelah. 1 The general situation, where CH fails, may be informed by the Kiesler-Shelah isomorphism theorem, which asserts that two first-order structures have isomorphic ultrapowers if and only if they have the same first-order theory. In particular, for any group$G$at all, of any size, we may take a countable elementary subgroup$H$, meaning in particular that they have the same first-order theory, and so there is a nonprincipal ultrafilter$U$on an index set$I$such that the ultrapowers$G^I/U\cong H^I/U$are isomorphic. Since every first-order structure maps elementarily into its ultrapowers, this means in particular that$G$maps elementarily (and hence monomorphically) into an ultrapower of$H$, a countable group. Thus, this fully answers the version of question 2 in which we allow the ultrafilter to live on a bigger index set (but still insist on each$G_i$being countable). If you want to insist that the ultrafilter concentrate on$\mathbb{N}$, then things become more complicated. If the CH holds, then the Kiesler-Shelah theorem shows that any two groups of size at most$2^{\aleph_0}$and with the same theory have isomorphic ultrapowers by an ultrafilter on$\aleph_0$, and so the desired result is attained. In the non-CH case, however, what we seem to get is that for any cardinal$\lambda$, if$\beta$is smallest such that$\lambda^\beta\gt\lambda$, then any two groups of size$\beta$with the same theory have isomorphic utrapowers using an ultrafilter on$\lambda\$. Thus, they each map into an ultrapower of the other.