The documentation simply got it wrong. It's old and probably nobody cared enough to fix it, assuming they spotted the mistake. The units don't work out, the possibility of a 0 in the denominator shouldn't be there either. There's no singularity.
According to Berkley SPICE 3f5 source code, the equation should read:
$$
\begin{aligned}
V_T(T) &= T \frac{k}{q(1{\rm\,V})} \\
I_S(T) &= I_S(T_0)
\left[\frac{T}{T_0}\right]^3
\exp \left[
\frac{E_g}{V_T(T)} \left(\frac{T}{T_0} - 1\right)
\right].
\end{aligned}
$$
The code involved, edited for readability, taken from src/lib/dev/bjt/bjttemp.c
, is:
vt = here->BJTtemp * CONSTKoverQ;
ratlog = log(here->BJTtemp/model->BJTtnom);
ratio1 = here->BJTtemp/model->BJTtnom -1;
factlog = ratio1 * model->BJTenergyGap/vt + model->BJTtempExpIS*ratlog;
factor = exp(factlog);
The 1V factor isn't written explicitly in the source code, but is required to get the units correct. Numerically it makes no difference. Multiplying by 1 is a no-op, not even worth putting in the source code. Back then, compilers might have had trouble optimizing such a multiplication-by-one out.
Reorganizing things a bit:
$$
\begin{aligned}
I_S(T) &= I_S(T_0)
\left[\frac{T}{T_0}\right]^3
\exp \left[
\frac{E_g q (1{\rm\,V})}{k T} \left(\frac{T}{T_0} - 1\right)
\right] \\
&= I_S(T_0)
\left[\frac{T}{T_0}\right]^3
\exp \left[
\frac{E_g q (1{\rm\,V})}{k} \left(\frac{1}{T_0} - \frac{1}{T}\right)
\right] \\
\end{aligned}
$$
Where, from src/include/const.h
and src/lib/dev/bjt/bjtsetup.c
we get:
$$
\begin{aligned}
q &= 1.6021918\cdot{10}^{-19}{\rm\,C} \\
k &= 1.3806226\cdot{10}^{-23}{\rm\,J\cdot K^{-1}} \\
E_g &= 1.11{\rm\,eV} \\
\end{aligned}
$$
This explains the "mysterious" \$E_g\cdot q(1{\rm\,V})\$ product in the formula: it converts \$E_g\$ customarily given in electron-Volts to Joules, to match the SI units of the Boltzmann constant value used.
In SI units, the formula becomes, simply:
$$
\begin{aligned}
I_S(T) &= I_S(T_0)
\left[\frac{T}{T_0}\right]^3
\exp \left[
\frac{E_g}{k} \left(\frac{1}{T_0} - \frac{1}{T}\right)
\right], \\
\end{aligned}
$$
just as jonk had commented.