For $A,B\in{\cal P}(\omega)$ we say $A\subseteq^* B$ if $A\setminus B$ is finite (that is, $A$ is "almost contained" in $B$). We write $A\simeq_{\text{fin}} B$ if $A\subseteq^* B$ and $B\subseteq^* A$ (that is, the sets $A, B$ are "almost the same set" except for finitely many elements). It is easy to see that $\simeq_{\text{fin}}$ is an equivalence relation on ${\cal P}(\omega)$.
We denote the collection of equivalence classes on ${\cal P}(\omega)$ with respect to $\simeq_{\text{fin}}$ by ${\cal P}(\omega)/(\text{fin})$. Using $\subseteq^*$ on representatives of equivalence classes, it is easy to see that we can make ${\cal P}(\omega)/(\text{fin})$ into a partially ordered set.
In the Noah Schweber's beautiful post inspiring this question, we learn that there is no order-embedding from $\omega_1$ into ${\cal P}(\omega)$, but $\omega_1$ can be order-embedded in ${\cal P}(\omega)/(\text{fin})$. So within the Continuum Hypothesis ${\sf (CH)}$ we get that $2^{\aleph_0} = \omega_1$ can be order-embedded in ${\cal P}(\omega)/(\text{fin})$.
Question. Can the cardinal $2^{\aleph_0}$ (well-ordered by $\in$) be order-embedded in ${\cal P}(\omega)/(\text{fin})$ even if $\neg{\sf(CH)}$? If not: can every member of $2^{\aleph_0}$ be order-embedded in ${\cal P}(\omega)/(\text{fin})$?