How is the vertical velocity defined in sigma-pressure-coordinates?
There is a misconception in your assumption:
$\omega = \frac{Dp}{Dt} = 0 $ means that the mass above the air parcel stays constant.

Proof: Assume quasistatic eqn:
-$\frac{1}{g}\, dp=\rho \, dz$;
apply operator of horizontal mean: $\overline{(\cdot)}= \frac{\int (\cdot)  \, dxdy}{\int dxdy}=\frac{\int (\cdot)  \, dA}{\Delta A}$
-$\frac{1}{g}\, \overline{dp} \, \Delta A=\rho \, dxdydz= \rho dV = dM$
Integrating between arbitrary pressure levels gives mass between those levels.

(1) You correctly estimated $\partial p_s/\partial y &lt;0 $ in order to have wind to the north, but
(2) you also set $\partial p_s/\partial t =0 $ in your derivation (to replace total derivative with advection of $p_s$). This means that the pressure field is constant in time.
Result
(1) and (2) yield to lower pressure in the north ($dp_s/dy&lt;0$) and decreasing surface pressure following the parcel ($dp_s/dt&lt;0$)
This means that an air parcel has to descend (in terms of $\sigma$) in order to satisfy the condition of constant mass lying over it ($\omega = \frac{Dp}{Dt} = 0 $):
$d\sigma/dt = a \, dp/dt- b\, dp_s/dt$; (with a,b const.)
=&gt; $d\sigma/dt &gt;0$
Example
If your parcel starts at 950 hPa, travels to north with falling surface pressure, condition $\omega=0$ means it stays at 950 hPa, but with (1) surface pressure drops to 950 hPa as the parcel continues north and the parcel miraculously is found to be at ground level.
Conclusion
As $\sigma$ depends on pressure and surface pressure, so does its time derivative.