Let $x(t)$ be your continuous-time bandlimited signal, modeled as a WSS random process with a PSD of $S_{xx}(\Omega)$ in the band $\Omega \in[-W,W]$, $\Omega$ in radians per second.

Sampling $x(t)$ at the critical rate $T_s = \frac{\pi}{W}$, yields the unquantized, discrete-time sequence $x[n] = x(nT_s)$. Let the associted PSD with $x[n]$ is
$$\Phi_{xx}(e^{j \omega}) = \frac{1}{T_s} S_{xx}(\frac{\omega}{T_s}) ~~~,~~~\omega \in [-\pi, \pi] \tag{1}$$

Assume some $N$ bits of quantization is used with a step-size $\Delta$. The process of quantization converts the unquantized sequence $x[n]$ into quantized (digital) $\hat{x}[n]$ with the quantization error denoted as $e[n]$ such that: $$\hat{x}[n] = x[n] + e[n] \tag{2}$$

For simplification reasons, the quantization error $e[n]$ is assumed to be an i.i.d., white-noise of independent (from $x[n]$ and itself at nonzero lags) WSS random process. Which, therefore has the PSD given by :

$$ S_{ee}(e^{j\omega}) = \frac{\Delta ^2}{12} ~~~,~~~\pi < \omega < \pi \tag{3}$$

Notice, how I modeled the quantization noise process, directly in discrete-time, with a power spectral density independent of sampling period $T_s$. This means that whatever sampling rate is used to obtain the quantized samples $\hat{x}[n]$, the assoiced noise power at the discrete-time will be the same as $\sigma_e^2 = \Delta^2/12$ in the range $-\pi < \omega < \pi$. It only depends on the number of bits; hence the step-size $\Delta$.

However, the same is not true for the PSD of the signal $x[n]$. As the sampling rate changes, its PSD will scale both in amplitude and in the frequency band as given by the relation. Assume we employ $M$ times oversampling so that $T' = T_s/M$, then:

$$ x[n] = x(n T_s) \longleftrightarrow \Phi_{xx}(e^{j\omega}) \implies \\ x_o[n]=x(n \frac{T_s}{M}) \longleftrightarrow \begin{cases}{ M \cdot \Phi_{xx}(e^{j\omega/M}) ~~~,~~~|\omega| < \frac{\pi}{M} \\ 0 ~~~,~~~ \frac{\pi}{M} < |\omega| < \pi }\end{cases} \tag{4}$$

Now, if you lowpass filter the quantized oversampled sequence $x_o[n]$ and decimate the result back into its critical rate, both PSDs of $x_o[n]$ and $e[n]$ will be reduced by $M$, however, this will only revert PSD of $x[n]$ into its original, while reducing the equivalent noise power of quantization into $1/M$ of its original value, thus enabling one to use more bits to represent the original signal samples $x(nT_s)$.

Let $x(t)$ be your continuous-time bandlimited signal, modeled as a WSS random process with a PSD of $S_{xx}(\Omega)$ in the band $\Omega \in[-W,W]$, $\Omega$ in radians per second.

Sampling $x(t)$ at the critical rate $T_s = \frac{\pi}{W}$, yields the unquantized, discrete-time sequence $x[n] = x(nT_s)$. Let the associted PSD with $x[n]$ is
$$\Phi_{xx}(e^{j \omega}) = \frac{1}{T_s} S_{xx}(\frac{\omega}{T_s}) ~~~,~~~\omega \in [-\pi, \pi] \tag{1}$$

Assume some $N$ bits of quantization is used with a step-size $\Delta$. The process of quantization converts the unquantized sequence $x[n]$ into quantized (digital) $\hat{x}[n]$ with the quantization error denoted as $e[n]$ such that: $$\hat{x}[n] = x[n] + e[n] \tag{2}$$

For simplification reasons, the quantization error $e[n]$ is assumed to be an i.i.d., white-noise of independent (from $x[n]$ and itself at nonzero lags) WSS random process uniformly distributed in $[-\frac{\Delta}{2} ~,~ \frac{\Delta}{2}]$. Which, therefore has the PSD given by :

$$ S_{ee}(e^{j\omega}) = \frac{\Delta ^2}{12} ~~~,~~~\pi < \omega < \pi \tag{3}$$

Notice, how I modeled the quantization noise process, directly in discrete-time, with a power spectral density independent of sampling period $T_s$. This means that whatever sampling rate is used to obtain the quantized samples $\hat{x}[n]$, the assoiced noise power at the discrete-time will be the same as $\sigma_e^2 = \Delta^2/12$ in the range $-\pi < \omega < \pi$. It only depends on the number of bits; hence the step-size $\Delta$.

However, the same is not true for the PSD of the signal $x[n]$. As the sampling rate changes, its PSD will scale both in amplitude and in the frequency band as given by the relation. Assume we employ $M$ times oversampling so that $T' = T_s/M$, then:

$$ x[n] = x(n T_s) \longleftrightarrow \Phi_{xx}(e^{j\omega}) \implies \\ x_o[n]=x(n \frac{T_s}{M}) \longleftrightarrow \begin{cases}{ M \cdot \Phi_{xx}(e^{j\omega/M}) ~~~,~~~|\omega| < \frac{\pi}{M} \\ 0 ~~~,~~~ \frac{\pi}{M} < |\omega| < \pi }\end{cases} \tag{4}$$

Now, if you lowpass filter the quantized oversampled sequence $x_o[n]$ and decimate the result back into its critical rate, both PSDs of $x_o[n]$ and $e[n]$ will be reduced by $M$, however, this will only revert PSD of $x[n]$ into its original, while reducing the equivalent noise power of quantization into $1/M$ of its original value, thus enabling one to use more bits to represent the original signal samples $x(nT_s)$.