Is the sampling period arbitrary for a bilinear transform, and why?
The sampling frequency is not arbitrary and important to specify for pre-warping. This is clear when reviewing how each analog frequency in the bilinear transform is mapped one-one to a digital frequency according to:
$$f = \frac{F_d}{F_s} =  \frac{1}{\pi}\tan^{-1} \bigg(\pi\frac{F_a}{F_s}\bigg) \tag{1} \label{1}$$
Where:
$f = F_d/F_s$ is the normalized digital frequency in units of cycles/sample,   $F_a$ is an analog frequency in Hz 
$F_s$ is the sampling rate in Hz 
Observe that $tan^{-1}(\infty) = \pi$ and thus the Bilinear Transform &quot;warps&quot; the entire analog frequency domain from $F=0$ to $F= \infty$ to the digital frequency domain of $f=0$ to $f=1$ (where $f=1$ is the sampling rate in units of normalized frequency), but the specific mapping of each analog frequency over that range is dependent on the value of $F_s$ that is used, as given by \ref{1}.
Normalizing the discrete-time frequency domain is commonly done such that the resulting spectrum associated with DC to the sampling rate extends from $f =0 \text{ to } 1$ cycles/sample, which may be the source of the OP's confusion.