How to derive entropy from density of states?

$$S=k \ln [\Omega(E)] = k \ln [\omega (E) \delta E] = k \ln [\omega(E)] +k \ln (\delta E)$$

Last term is an arbitrary constant, so that we can set

$$S = k \ln[\omega (E)]$$

from which

$$\delta S = k \frac{\delta \omega}{\omega}$$

If we can ignore the power contribution and set $\omega (E) \simeq e^{\beta E}$, we get

$$\delta S = k \frac{\delta(e^{\beta E})}{e^{\beta E}} = k \frac{\beta \ e^{\beta E} \delta E}{e^{\beta E}} = k \ \beta \delta E$$

More about entropy and density of states: here.

As @valerio92 points out, your mistake is that $S = k \ln (\omega\, \delta E)$, not $\delta S$. To get $\delta S$, you differentiate the right-hand expression to get $\delta S = k \frac{\delta \omega}{\omega}$, and the $\delta E$ drops out and you get an expression with the right dimensions. The notation is a bit misleading, because the $\delta$ in the $\delta E$ is not a differential corresponding to the $\delta$ in the $\delta S$ - it just denotes that we should think of $\delta E$ as a small constant quantity. Once you differentiate the expression for $S$, the "differential" $\delta$ actually ends up on the $\omega$, which is the actual variable quantity.