DC Free Transmission SNR
Noise performance depends on the difference between levels. Since both your signals have a difference of 1 volt, they have the same resistance to noise.

However, they have different energy. Let's say that your data signal has a period of 1 second. Then, the pulses you transmit have the following energy:


The 0V pulse has energy 0 J.
The 1V pulse has energy 1 J.


Then, on average your signal requires 0.5 J per pulse.

The second signal uses pulses of the following energy:


The 0.5 V pulse has energy 0.25 J.
The -0.5 V pulse has energy 0.25 J.


So, on average the energy in this signal is 0.25 J per pulse.

So, removing DC results in a signal with the same error performance but less energy. This is equivalent to saying that you have a larger SNR.

(Note that I have made an assumption that is common in communications, that the load resistance is \$R_L = 1\,\Omega\$. In any case, when comparing the SNR of the two signals, \$R_L\$ cancels out, so its actual value is not important).