The answer lies with induction.
The oxygen is better able to stabilize the residual negative charge on the conjugate base of methanal by virtue of residing closer to the negative formal charge.
On the conjugate base of ethanal note that the electronegative oxygen is one carbon removed from the negative formal charge while on the conjugate base of methanal, the oxygen is on the same carbon with the negative formal charge. Thus, the oxygen in methanal will be better able to withdraw electron density from the spot where heterolytic bond clevage occurred. This follows from Coulomb's Law:
${F=k\frac{q_1q_2}{r^2}}$
As we can see from Coulomb's law, force and distance are related in an inverse-square fashion. In other words, increasing the distance of the oxygen nucleus from the site of the negative charge dramatically decreases the electron-withdrawing effects of oxygen.