If the continuum hypothesis holds, then this is impossible. To see this, consider a family of $\omega_2$ many pairwise disjoint subsets of $\omega_2$, and for each $x$ in this family, consider the range of $e_x$, which is a subset of $\omega$. By the continuum hypothesis, there are only $\omega_1$ many possible ranges, and so there must be two disjoint countable subsets $x,y\subset\omega_2$ with $\text{ran}(e_x)=\text{ran}(e_y)$. Let $z=x\cup y$ and consider $e_z$. It cannot be that $e_z$ has only finite difference from $e_x$ and also from $e_y$, if $e_z$ is injective.

I'm not yet sure what happens when CH fails...

Meanwhile, here are a few slightly weaker generalizations that are possible.

There is a family of injective functions $e_x:x\to \omega$ for all countable $x\subset\omega_1$, that is, for $x\in [\omega_1]^\omega$, such that $e_x$ and $e_y$ have at most finitely many disagreements on $x\cap y$. To see this, first fix a family of functions $e_\alpha$ for $\alpha\lt\omega_1$ as mentioned at the outset of the question, so that $e_\alpha$ injects $\alpha$ to $\omega$, and any two of them disagree only finitely. The point is that we may extend this family to $e_x:x\to \omega$ for $x\in[\omega_1]^\omega$ by defining $e_x=e_\alpha\upharpoonright x$, where $\alpha=\sup(x)$. The point is that if $\sup(y)=\beta$, then $e_\beta$ and $e_\alpha$ have only finitely many disagreements on their common domain, and so also $e_x$ and $e_y$ will therefore also have only finitely many disagreements.

There is a family of injective functions $f_x:x\to \omega$ for all $x\in(\omega_2)^\omega$, meaning $x\subset\omega_2$ and $x$ has order type $\omega$, such that $f_x$ and $f_y$ have at most finitely many disagreements on $x\cap y$. To see this, fix the $e_x$ as in the previous paragraph, and also for each $\beta\lt\omega_2$, fix an injection $\pi_\beta:\beta\to \omega_1$, and now for $x\subset\omega_2$ of order type $\omega$, let $\beta=\sup(x)$, and consider $\pi_\beta[x]\subset \omega_1$. We may let $f_x=(e_{\pi_\beta[x]}\circ\pi_\beta)\upharpoonright x$. That is, we use $\sup(x)$ to tell us how to regard $x$ as really a subset of $\omega_1$, and then we apply the corresponding function from before. The point is that if $x$ and $y$ are two increasing $\omega$ sequences in $\omega_2$, then if they have different suprema, then $x\cap y$ is finite anyway and we needn't worry about it, and if they have the same suprema, then $f_x$ and $f_y$ agree completely.