Nice question! I like it very much.

Sure, we can do this. We'll also arrange that
$\text{ran}(f_\alpha)$ is coinfinite. That will make the successor
steps easy, since there is always another point available. The
difficulty is what to do at limits. Suppose we have $f_\alpha$ for
$\alpha<\lambda$, where $\lambda$ is a countable limit ordinal. We want to define the function $f_\lambda$. In order to do this, pick an increasing cofinal sequence
$\alpha_0<\alpha_1<\alpha_2<\cdots<\lambda$ with
$\sup_n\alpha_n=\lambda$. We shall define $f_\lambda$ in blocks. First, let $f_\lambda$ agree with
$f_{\alpha_0}$ up to $\alpha_0$. Next, on the interval
$[\alpha_0,\alpha_1)$, we make $f_\lambda$ agree with
$f_{\alpha_1}$, except modified so that it is still injective in combination with what we already did below $\alpha_0$.
This modification will require only at most finitely many changes to $f_{\alpha_1}\upharpoonright[\alpha_0,\alpha_1)$, since
$f_{\alpha_0}$ and $f_{\alpha_1}$ agree except on a finite set.
Continuing, we define $f_\lambda$ on the interval
$[\alpha_n,\alpha_{n+1})$ to agree with $f_{\alpha_{n+1}}$, except
fixed up again to be injective, which will require only finitely many
changes. Ultimately, in this way we'll get an injective function
$f_\lambda:\lambda\to\omega$. For any $\alpha<\lambda$, we have
$\alpha<\alpha_n$ for some $n$, and consequently
$f_\lambda\upharpoonright\alpha$ was obtained from finite
modifications of the $f_{\alpha_k}$'s for $k<n$, each of which had
only finite difference from $f_\alpha$ up to $\alpha$. And so our
function $f_\lambda$ differs from $f_\alpha$ on $\alpha$ only at
most finitely. Lastly, to ensure that $f_\lambda$ has coinfinite
range, we may if necessary simply change the values on the
ordinals $\alpha_n$, since this will involve only finitely many
changes below any given $\alpha<\lambda$. Thus, we have constructed
the function $f_\lambda$ with the desired properties, the
construction may proceed up to $\omega_1$.