Write it as
$$y=Ae^{-bt}e^{-\sqrt{b^2-w^2}t}+Be^{-bt}e^{+\sqrt{b^2-w^2}t}$$
$$y=e^{-bt}(Ae^{-\sqrt{b^2-w^2}t}+Be^{+\sqrt{b^2-w^2}t})$$
This looks promising. Because the exponent tends toward $0$ in the limit, you can use Taylor expansion:
$$y=e^{-bt}(A-A\sqrt{b^2-w^2}t+\frac{A}{2}(\sqrt{b^2-w^2}t)^2+\cdots+$$
$$+B+B\sqrt{b^2-w^2}t+\frac{B}{2}(\sqrt{b^2-w^2}t)^2+\cdots)$$
$$y=e^{-bt}((A+B)-(A-B)\sqrt{b^2-w^2}t+\frac12(A+B)(b^2-w^2)o(t^2))$$
The important part to realize is, that you can't treat $A$ and $B$ as constants during the limit. The solution of the DE depends on the initial conditions, so the $A$ and $B$ for fixed boundary conditions depend on $b$ and $w$. You can do this explicitly by yourself. Just invent some initial condition $u=y(0)$ and $v=y'(0)$ and solve it, then take the limit.
However, simple counting of degrees of freedom above tells you what happens. You can always choose $A$ and $B$ such that $A-B$ tends to infinity in a way that cancels out the $\sqrt{b^2-w^2}$ in the limit. That's exactly what happens if you fix the initial condition. However, the quadratic term has the same $A+B$ behaviour as the constant term... so, this one has to go to $0$ (otherwise $A+B$ would need to diverge, and that makes the entire solution divergent).
You can also write the initial solution as a combination of hyperbolic sine and cosine. There it is more obvious that $\cosh(x)\to 1$ and $\sinh(x)\to x$ for small arguments. This also works from the other side, if $b$ goes to $w$ from below (you get $\cos(x)\to 1$ and $\sin(x)\to x$ then).
Edit:
The limiting process would go as such. Let $y(0)=u$ and $y'(0)=v$. These mean
$$y(0)=u=A+B$$
$$y'(0)=v=-\lambda A-\mu B=-\frac12(\lambda+\mu)(A+B)-\frac12(\lambda-\mu)(A-B)$$
This seems like black magic, but it's just a shorter way instead of expressing $A$ and $B$ with $u$ and $v$.
So, express:
$$A+B=u$$
$$A-B=\frac{-v-\frac12u(\lambda+\mu)}{\frac12(\lambda-\mu)}=\frac{-v-bu}{\sqrt{b^2-w^2}}$$
The Taylor expanded solution becomes:
$$y=e^{-bt}((A+B)-(A-B)\sqrt{b^2-w^2}t+\frac12(A+B)(b^2-w^2)o(t^2))=$$
$$=e^{-bt}(u-\frac{-v-bu}{\sqrt{b^2-w^2}}\sqrt{b^2-w^2}t+\frac12u(b^2-w^2)o(t^2))$$
$$=e^{-bt}(u+(v+bu)t+\frac12u(b^2-w^2)o(t^2))$$
You see, now if you take the limit $b\to w$, the linear term survives.
The characteristic polynomial is $aX^2+bX+c$ where $a,b,c>0$. For $r \geq 0$, $ar^2+br+c>0$ so any real root is negative. This argument works even in the higher degree case.
Best Answer
When you write the solution of the overdamped system as $$ \eqalign{ & f(t) = c_{\,1} e^{\,\rho \,t + \omega \,t} + c_{\,2} e^{\,\rho \,t - \omega \,t} = \left( {c_{\,1} e^{\,\omega \,t} + c_{\,2} e^{\, - \omega \,t} } \right)e^{\,\rho \,t} = \cr & = \left( {a\cosh \,\left( {\omega \,t} \right) + b\sinh \,\left( {\omega \,t} \right)} \right)e^{\,\rho \,t} \cr} $$
and impose the initial conditions, for instance for $f(0)$ and $f'(0)$, you get $$ \left\{ \matrix{ f(0) = a \hfill \cr f'(0) = \,\,\rho a + b\omega \quad \Rightarrow \quad b = \;{1 \over w}\left( {f'(0) - \,\,\rho f(0)} \right) \hfill \cr} \right. $$
so $$ f(t) = \left( {f(0)\cosh \,\left( {\omega \,t} \right) + {1 \over w}\left( {f'(0) - \,\,\rho f(0)} \right)\sinh \,\left( {\omega \,t} \right)} \right)e^{\,\rho \,t} $$
Now, if the damping approaches the critical value, that is $\omega \to 0$, then
$$ \bbox[lightyellow] { \mathop {\lim }\limits_{\omega \, \to \,0} f(t) = \left( {f(0) + \left( {f'(0) - \,\,\rho f(0)} \right)t} \right)e^{\,\rho \,t} }$$
And similarly, starting from an under-damped system, where you have normal $\sin $ and $\cos$ instead of the hyperbolic version.
So the critically-damped response is at the frontier between the two, mathematically and physically, and not easy distinguishable at first sight when very near to the critical value.
In fact, the under-damping case will always be evidenced by more or less visible oscillations.
In case of overdamping instead, since for passive system $\rho$ is negative, the exponential decay is prevalent and masking in the long time. In the short time instead, the difference between $\cosh (\omega t)$ and $1$ and $\sinh (\omega t)$ and $\omega t$ is not appreciable.