A question about the exact formula for the collector current
As per Ebers–Moll Model of an NPN transistor:

$$I_E = I_{ES} (e^{V_{BE}/V_T} -1) -\alpha_RI_{CS}(e^{V_{BC}/V_T}-1)$$
$$I_C = \alpha_FI_{ES} (e^{V_{BE}/V_T} -1) -I_{CS}(e^{V_{BC}/V_T}-1)$$
where:

\$ I_{ES}\$ = reverse saturation current of Base-Emitter Junction.

\$ \alpha_R\$ = reverse current gain

\$ \alpha_F\$ = foward current gain

\$ I_{CS}\$ =  reverse saturation current of Base-Collector Junction.

Approximated Ebers-Moll Models look like:
$$I_E \approx I_{ES} (e^{V_{BE}/V_T} -1) $$
$$I_C \approx \alpha_FI_{ES} (e^{V_{BE}/V_T} -1) $$
which satisfies the normal convention: \$ I_C = \alpha_F I_E \$

Therefore, whatever source you took the figure from, By \$ I_S \$ they mean, \$ \alpha_F I_{ES} \$.

If you are looking for where is \$ I_{CBO} \$, it is there in the original equation of \$ I_C \$ , the term --> \$-I_{CS}(e^{V_{BC}/V_T}-1) \$ is nothing but equals to \$ + I_{CBO}\$.

From the original Ebers-Moll Model equations, you can see that  even \$ I_C = \alpha_F I_E + I_{CBO} \$  is an approximation, as it neglects the second term in the expression for \$I_E\$.