Martin is right. It is very simple to see it: just consider for a moment to add this extra superpotential term $$ \delta W(\Phi_i)=\frac{1}{2}m_{i j}\Phi_i \Phi_j $$ where $\Phi_i$ are chiral superfields. This extra $W$ corresponds to an extra lagrangian term $$ \delta\mathcal{L}=-\frac{1}{2}(\psi_i\psi_j+h.c)- \bar{\phi}_iM^2_{ij}\phi_j\qquad M^2_{ij}=\bar{m}_{ik}m_{jk}\,. $$ Therefore, starting with $W$ and adding a soft mass for the fermions $\delta\mathcal{L}_{\mathrm{soft}-\psi}=-\frac{1}{2}(\psi_i\psi_j+h.c)$ $$ W(\Phi)\rightarrow \mathcal{L}=\mathcal{L}_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\psi} $$ is just the same as doing starting with $W+\delta W$ (with $\delta W$ in the first equation above) and add a soft mass terms for the scalars, $\delta\mathcal{L}_{\mathrm{soft}-\phi}=+\bar{\phi}_iM^2_{ij}\phi_j$ $$ W(\Phi)+\delta W(\Phi)\rightarrow \mathcal{L}=\mathcal{L}^\prime_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\phi}=\mathcal{L}_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\psi} $$
In other words, it is intuitive that only the mass splitting between fermions and boson that matters and should be considered. Adding soft masses that respect susy, that is that do not split fermions and bosons, it is just like adding a mass term to $W$