One of the reasons involve the changing of the transfer function. Considering \$Z_f\$ as the feedback equivalent impedance and \$Z_i\$ as the input equivalent impedance, the transfer function for an inverting opamp is \$-Z_f/Z_i\$, while for a non-inverting opamp, \$Z_i\$ is the grounded equivalent impedance, and the gain is \$1+Z_f/Z_i\$.
Now let's consider a simple lowpass, \$\omega/(s+\omega)\$. If you were to use a capacitor (\$C_1\$) in parallel with a resistor (\$R_1\$) as \$Z_f\$, and a resistor (\$R_2\$) as \$Z_i\$, you'd have this resultant transfer function:
$$
H(s)=-\frac{C_1||R_1}{R_2}=-\frac{\frac{1}{sC_1+\frac{1}{R_1}}}{R_2}=-\frac{R_1}{R_1R_2Cs+R_2}=-\frac{R_1}{R_2}\frac{\frac{1}{R_1C}}{s+\frac{1}{R_1C}}
$$
Which is the lowpass transfer function with a gain. If you apply the same elements to the non-inverting configuration, this is what you get:
$$
H(s)=\left(1+\frac{R_1}{R_2}\right)\frac{\frac{1}{R_1C}}{s+\frac{1}{R_1C}}=\left(1+\frac{R_1}{R_2}\right)\frac{\frac{R_2}{R_1+R_2}s+\frac{1}{R_1C}}{s+\frac{1}{R_1C}}
$$
Which is a shelf-lowpass, of the form \$(s+\omega_z)/(s+\omega_p)\$. That \$1\$ changes everything: it adds the denominator to the numerator.
What you have there is a high-pass, as an inverting configuration, which transforms into a shelf-highpass for the non-inverting configuration. The reason why you don't see this response in your picture is because you're using unipolar supply for the inverting opamp, while using bipolar supply for the non-inverting one. If you'll make both opamps suplied from the same sources, the results should change.
