There's a difference between the saturation current \$I_S\$ as defined for a bipolar junction transistor model, and saturation current \$I_S\$ in the context of a diode. If the base-emitter junction is modelled as a diode, which obeys the diode equation, then the equation for base current through it, \$I_B\$, would be written:
$$ I_B = I_{ES}\left(e^{\frac{V}{\eta V_T}}-1 \right) $$
Here, \$I_{ES}\$ is the base-emitter junction's saturation current, as distinct from saturation current parameter \$I_S\$ used in transistor models. The two are related:
$$ I_S = \beta I_{ES} $$
Substituting \$I_{ES}\$ into your equation gives:
$$
\begin{aligned}
I_C &\approx I_Se^{\frac{V}{V_T}} \\ \\
&\approx \beta I_{ES}e^{\frac{V}{V_T}}
\end{aligned}
$$
\$\beta\$ didn't disappear, it just got hidden, or encapsulated, inside the model's \$I_S\$ parameter, in the same way temperature \$T\$ is hidden inside \$V_T\$. I've expanded on this at the end, if you're interested.
Your first circuit calculates \$I_B\$ correctly, given the assumption that a diode develops \$V_{BE}\approx 0.7V\$. That assumption permits you to estimate base current with some accuracy. The second circuit though, which employs a resistor voltage divider to bias the transistor, is dubious, and not as trivial as it seems.
The base-emitter junction's "clamping" of \$V_{BE}\$ to a maximum near 0.7V means that the following claim is questionable:
$$ V_{BE} = V_{CC}\frac{R_2}{R_1 + R_2} $$
It would be true if there were not a diode across R2, but there is a diode, the B-E junction. That diode will oppose (to some extent) any rise in voltage across R2 beyond 0.6V or so, so that last equation is therefore only true for:
$$ V_{CC}\frac{R_2}{R_1 + R_2} < 0.6V $$
Above that threshold, you begin to get both base current and current through R2, which sum according to KCL, and the equation no longer accurately models circuit state. I don't see any analogue application in which this "divider" method of biasing is of any use. You might see this used in a digital/switching context, where there's no quiescent state to obtain and maintain.
The only way to obtain an accurate value for \$I_B\$ in the second circuit is to solve the set of simultaneous equations derived from applications of KVL, KCL and Ohm's laws:
simulate this circuit – Schematic created using CircuitLab
$$ I_1 = I_2 + I_B $$
$$ I_1 = \frac{V_{CC} - V_B}{R_1} $$
$$ I_2 = \frac{V_B}{R_2} $$
$$ I_B = \frac{I_S}{\beta} e^{\frac{V_B}{V_T}} $$
I'll also add that both methods of biasing a BJT (for an analogue application) are bad, mainly because thermal voltage \$V_T\$ is temperature dependent, and collector current \$I_C\$ will vary significantly with even small variations in temperature.
Mostly you find that biasing is achieved via negative feedback, which can be done in many ways, such as a resistance between collector and base, or via emitter degeneration.
The diode equation:
$$ I = I_S\left(e^{\frac{V}{\eta V_T}}-1 \right) $$
The transistor's base-emitter junction is a diode, obeying that relationship. Compared to typical discrete diodes, B-E junctions tend to have very good "ideality", \$\eta =1\$.
I will replace \$I_S\$ in the standard diode equation with \$I_{ES}\$, in keeping with the convention used in this Wikipedia article about the BJT, indicating that it refers to the base-emitter junction's saturation current, as distinct from parameter \$I_S\$ in the transistor's simplified model.
$$
\begin{aligned}
I_B &= I_{ES}\left(e^{\frac{V_{BE}}{\eta V_T}}-1 \right) \\ \\
I_C &= \beta I_B \\ \\
&= \beta I_{ES}\left(e^{\frac{V_{BE}}{V_T}}-1 \right) \\ \\
\end{aligned}
$$
Thermal voltage is \$V_T\approx 25mV\$, and usually the transistor is operated with \$V_{BE}\$ far exceeding that, over 0.6V or so. Since \$V_{BE} >> V_T\$ we can make the approximation:
$$ e^{\frac{V_{BE}}{V_T}}-1 \approx e^{\frac{V_{BE}}{V_T}} $$
The equation then becomes:
$$ I_C = \beta I_{ES}e^{\frac{V_{BE}}{V_T}} $$
Encapsulate the term \$\beta I_{ES}\$ into a single parameter \$I_S\$ to obtain a scaled saturation current for use in simplified transistor models:
$$
\begin{aligned}
I_S &= \beta I_{ES} \\ \\
I_C &= I_Se^{\frac{V_{BE}}{V_T}}
\end{aligned}
$$
