= Solution
First prove the <torsion in an HNN extension> statement directly from <Britton's lemma>. Let $z$ have finite order in an <HNN extension> of $G$. Among its conjugates, choose a reduced normal-form word with the smallest possible number $k$ of <stable letters>. If $k=0$, that conjugate lies in the embedded base $G$, so its finite order and the base's <torsion-free group> property force it to be the identity.
Suppose instead that $k>0$. There are no internal <pinches in an HNN extension> because the word is reduced. There is also no pinch at the join of its end with its beginning. Indeed, write it as $g_0t^{\varepsilon_1}g_1\cdots t^{\varepsilon_k}g_k$ and conjugate off $g_0$ to obtain $t^{\varepsilon_1}g_1\cdots t^{\varepsilon_k}(g_kg_0)$. If the joined subword $t^{\varepsilon_k}(g_kg_0)t^{\varepsilon_1}$ were a pinch, its two signs would be opposite; conjugating by $t^{\varepsilon_1}$ and using the pinch relation would reduce the stable-letter count by two, contradicting minimality. Thus every positive power is reduced with exactly $rk$ stable letters in the $r$th power. <Britton's lemma> makes every such power nonidentity, contradicting finite order. We have proved
$$
\boxed{G\text{ torsion-free}\quad\Longrightarrow\quad\text{every HNN extension of }G\text{ is torsion-free}.}
$$
The base embeds by the same lemma, so finite order has not been lost or artificially introduced by a quotient.
For the <countable torsion-free embedding with two conjugacy classes>, start with $G_0=G$. Given a countable torsion-free $G_j$, list all ordered pairs $(a,b)$ of its nonidentity elements. Both elements have infinite order, so the assignment $a^r\mapsto b^r$ is an isomorphism between their infinite <cyclic subgroups>. Successively adjoin one stable letter for each pair with relation $t_{a,b}^{-1}at_{a,b}=b$. Each <HNN extension> embeds the previous group and is torsion-free by the result just proved. Let $G_{j+1}$ be the union after all pairs from $G_j$ have been treated. It remains countable because only countably many generators were added, and torsion-free because any element and any purported finite-order equation occur in a finite stage.
Repeat this whole process: new stable letters and their products must also be included in later pair lists. In the final increasing union
$$
\boxed{\widehat G=\bigcup_{j\geq0}G_j,}
$$
the original $G$ embeds and torsion-freeness and countability persist. Any two nonidentity elements lie together in some $G_j$ and are conjugate in $G_{j+1}$. Thus \b[every countable torsion-free group embeds in a countable torsion-free group in which all nonidentity elements are conjugate]. If the initial group is nontrivial there are exactly two <conjugacy classes>; for the trivial group the stated condition is vacuous, or one can begin with $\mathbb Z$ instead. The listing is set-theoretic; no algorithm for recognizing nonidentity elements is assumed.
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