This is just as it turns out when you do the math. The discrete-time Fourier transform (DTFT) of the sampled continuous-time impulse response $h(t)$ is
$$H_d(e^{j\omega T})=\sum_nh(nT)e^{-jn\omega T}\tag{1}$$
With
$$h(nT)e^{-jn\omega T}=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}\delta(t-nT)dt\tag{2}$$
this can be written as
$$\begin{align}H_d(e^{j\omega T})&=\sum_n\int_{-\infty}^{\infty}h(t)e^{-j\omega t}\delta(t-nT)dt\\&=\int_{-\infty}^{\infty}\left[h(t)\sum_n\delta(t-nT)\right]e^{-j\omega t}dt\\&=\mathcal{F}\left\{h(t)\sum_n\delta(t-nT)\right\}\\&=\frac{1}{2\pi}H(\omega)\star\frac{2\pi}{T}\sum_k\delta\left(\omega-\frac{2\pi k}{T}\right)\\&=\frac{1}{T}\sum_kH\left(\omega-\frac{2\pi k}{T}\right)\tag{3}\end{align}$$
where $H(\omega)$ is the Fourier transform of $h(t)$, and $\star$ denotes convolution. From $(3)$ we see that the DTFT of the sampled impulse response equals the sum of shifted spectra of $h(t)$, scaled by $1/T$.
If we assume that $H(\omega)$ is approximately band-limited and that $T$ is chosen sufficiently small such that aliasing becomes negligible, we obtain the approximation
$$H_d(e^{j\omega T})\approx\frac{1}{T}H(\omega),\qquad |\omega|<\frac{\pi}{T}\tag{4}$$
For the step-invariance method, we use samples of the step response instead of samples of the impulse response, and we obtain a relation analogous to $(3)$ between the DTFT $G_d(e^{j\omega T})$ of the step response of the discrete-time system, and the Fourier transform $G(\omega)$ of the continuous-time step response:
$$G_d(e^{j\omega T})=\frac{1}{T}\sum_kG\left(\omega-\frac{2\pi k}{T}\right)\tag{5}$$
In order to obtain the frequency response $H_d(e^{j\omega T})$ we multiply $(5)$ by $1-e^{-j\omega T}$, because the impulse response is obtained by computing a first-order difference of the step response:
$$H_d(e^{j\omega T})=\left(1-e^{-j\omega T}\right)G_d(e^{j\omega T})=\frac{1-e^{-j\omega T}}{T}\sum_kG\left(\omega-\frac{2\pi k}{T}\right)\tag{6}$$
For frequencies that are small compared to the sampling frequency, i.e., for $|\omega T|\ll 1$ we obtain from $(6)$
$$\begin{align}H_d(e^{j\omega T})&\approx\frac{1-(1-j\omega T)}{T}\sum_kG\left(\omega-\frac{2\pi k}{T}\right)\\&=j\omega \sum_kG\left(\omega-\frac{2\pi k}{T}\right)\tag{7}\end{align}$$
If we again assume that aliasing can be neglected, we arrive at
$$H_d(e^{j\omega T})\approx j\omega G(\omega)=H(\omega),\qquad |\omega|<\frac{\pi}{T}\tag{8}$$
From $(8)$ we see that, unlike for the impulse invariance method, the step invariance method doesn't involve a scaling of the continuous-time frequency response.
