Relation between erfc and the BER Antipodal Signalling has a bit error probability of $P_b = \frac{1}{2} \mathrm{erfc}\big(\sqrt{\frac{E_s}{N_0}}\big)$ and on-off-signalling has a bit error probability of $\frac{1}{2} \mathrm{erfc}\big(\sqrt{\frac{E_s}{2 N_0}}\big)$. Why does this correspond to a higher SNR of 3 dB which we need for on-off-signalling to achieve the same bit error probability as with antipodal signalling?

I know that the factor 2 corresponds to 3 dB. But the complementary error function does not decrease by a factor of 2 but by a factor of $\sqrt{2}$
Assuming your bit error probability formulas are correct, for antipodal signalling

$$P_{b, aS} = \frac{1}{2} \mathrm{erfc}\left(\sqrt{\frac{E_{s,aS}}{N_0}}\right)$$

and for on-off signalling

$$P_{b, oS} = \frac{1}{2} \mathrm{erfc}\left(\sqrt{\frac{E_{s,oS}}{2N_0}}\right)$$

If these two error probability are identical, i.e. $P_{b, aS} = P_{b, oS}$,

$$\frac{1}{2} \mathrm{erfc}\left(\sqrt{\frac{E_{s,aS}}{N_0}}\right) = \frac{1}{2} \mathrm{erfc}\left(\sqrt{\frac{E_{s,oS}}{2N_0}}\right) \tag{1}$$

Because both $\mathrm{erfc(.)}$ and $\sqrt{.}$ are bijective functions, $(1)$ implies $\frac{E_{s,oS}}{N_0} = 2\frac{E_{s,aS}}{N_0}$ in linear scale and 3dB difference in log scale.