the non-inverting terminal is fixed at ground, then it is impossible to have a differential input for the op-amp

Maybe the way you are thinking about "two perfect differential signals" is causing some confusion. Consider you have two different voltages \$V_A\$ and \$V_B\$, hence a differential signal. \$V_{DC} = (V_A + V_B)/2\$ is your DC component and \$V_{dif} = (V_A - V_B)\$ the difference. For convenience and to simplify the analysis by superposition, as you wrote in your question, you superimpose the signals \$V_{DC} \pm V_{dif} / 2\$.

If \$V_A = 0\$ and \$V_B = 1\$ than \$V_{DC} = 0.5\$ and \$V_{AC} = 0.5\$.

the non-inverting terminal is fixed at ground, then it is impossible to have a differential input for the op-amp

Maybe the way you are thinking about "two perfect differential signals" is causing some confusion. Consider you have two different voltages \$V_A\$ and \$V_B\$, hence a differential signal. \$V_{DC} = (V_A + V_B)/2\$ is your dc component and \$V_{dif} = (V_A - V_B)\$ the difference. For convenience and to simplify the analysis by superposition, as you wrote in your question, you superimpose the signals \$V_{DC} \pm V_{dif} / 2\$.

If \$V_A = 0\,\mathrm{V}\$ and \$V_B = 1\,\mathrm{V}\$ than \$V_{DC} = 0.5\,\mathrm{V}\$ and \$V_{AC} = 0.5\,\mathrm{V}\$.