The transistor going into saturation isn't a property of the transistor itself, but instead a property of the circuit surrounding the transistor and the transistor, as part of it.
The simplest case to imagine is an NPN switch. I'll present two different such switch circuits to make the above point concretely clear:
simulate this circuit – Schematic created using CircuitLab
Let's assume a perfect \$\beta=100\$ in both cases shown. We are going to increase the current source's current into the base (of either or both NPNs) from \$0\:\mu\textrm{A}\$ to \$100\:\mu\textrm{A}\$. In both cases, the relationship of \$I_C=\beta\cdot I_B=100\cdot I_B\$ will be considered to hold until some other limitation forces this relationship to change.
In the left circuit, as the base current rises, the permitted collector current also rises. When \$I_B=90\:\mu\textrm{A}\$, then \$I_C=9\:\textrm{mA}\$ and the voltage drop across \$R_1\$ will be \$I_C\cdot R_1=9\:\textrm{mA}\cdot 1\:\textrm{k}\Omega=9\:\textrm{V}\$. This works fine, as the collector voltage will then be \$V_C=10\:\textrm{V}-I_C\cdot R_1=1\:\textrm{V}\$. This voltage is between the ground rail of \$0\:\textrm{V}\$ and the power supply rail of \$10\:\textrm{V}\$ and since the transistor isn't yet in saturation, as I still expect \$V_C\ge V_B\$ (given \$V_E=0\:\textrm{V}\$.) But as \$I_B\$ increases still further towards \$I_B=100\:\mu\textrm{A}\$, I would expect the collector voltage to drop still further to a final case where \$V_C=0\:\textrm{V}\$ as the voltage drop across \$R_1\$ reaches a full \$10\:\textrm{V}\$. This assumes that the collector actually can achieve such a situation. But it can't. But before discussing why, let's move to the right side.
Looking at the right side schematic, you see the same thing except that \$R_2=10\:\textrm{k}\Omega\$, instead. Other details remain the same. In this case, though, the voltage drop across \$R_2\$ will reach \$9\:\textrm{V}\$ when \$I_B=9\:\mu\textrm{A}\$ and \$I_C=900\:\mu\textrm{A}\$ and a full \$10\:\textrm{V}\$ voltage drop when \$I_B=10\:\mu\textrm{A}\$. Assuming this is even possible, what should then happen as the \$I_2\$ current source increases still further?
Well... nothing more can happen. There's no possible way for a still larger voltage drop across the collector resistor, \$R_2\$. To do so would require \$Q_2\$'s collector to move into a negative voltage value with respect to ground. But there are no sources for that negative voltage available and while \$Q_2\$'s guts might be able to produce voltages in between \$0\:\textrm{V}\$ and \$10\:\textrm{V}\$, \$Q_2\$'s guts can't manufacture voltages outside that range out of thin air. It just doesn't happen.
So the process stops here. More base current achieves nothing. You can apply it, of course. There's nothing to stop \$I_2\$ from continuing right on up to a full \$100\:\mu\textrm{A}\$. So that works just fine. But the collector voltage just cannot continue it's downward direction any more. So the collector current just stops, regardless of the base current. The result is that the effective \$\beta\$ drops from 100 to some lower value, then.
All this said, though, a real BJT can't even cause the collector voltage to match up exactly with its emitter voltage. The base-collector diode can go into a forward biased mode in order to allow the collector to drop. And it has to do that, if it is going to squeeze out the last remaining additional dribbles of collector current so that the voltage drop across the collector resistor can rise just a little bit more. But at some point before the collector voltage reaches the emitter voltage, the process halts. There must be at least a small voltage difference remaining, just to operate at all. This might cause the base-emitter diode to be forward biased with \$800\:\textrm{mV}\$ while the base-collector diode is forward biased with \$600\:\textrm{mV}\$, so that \$V_{CE}=200\:\textrm{mV}\$. But the base-collector diode cannot be more forward biased than the base-emitter diode. Because doing so would require the BJT to present an impossible collector voltage that it cannot observe and cannot just create out of thin air. (At least, in the circuits I've shown above.)
At this point, it should also be clear that the external circuit matters. These two circuits were identical except for the collector load. But the limitation of collector current depends upon the collector resistor value, as well as the BJT. So saturation is best not seen as just an internal detail of the BJT but instead that it also depends upon what is surrounding the BJT.
Another way of saying this is that the transistor enters gradually into saturation as the collector current, coupled with the collector load external to it, causes the collector voltage to move in such a way that the base-collector diode transitions from being reverse-biased into becoming forward-biased. While the BC junction is still reverse-biased, the transistor is in active mode. Once the BC junction transitions into forward-biased, the BJT is in saturated mode. However, saturation is gradual in the sense that \$\beta\$ gradually declines and doesn't suddenly change (it's not a switch-like effect) -- in the above examples where the base current is gradually changed.
For design purposes, if you want a switch behavior then you anticipate the above process and just design around some value of \$\beta\$ that you want to achieve for your switch. If you look it up on a datasheet, there will usually be a curve showing just how small of a difference between the collector and emitter is achievable given some desired \$\beta\$. Or, at least, an example for \$\beta=10\$ which is usually considered to be a highly saturated case for most (but not all) BJTs. Since the external circuit can be designed to force a low \$\beta\$ result, that all works fine. (Of course, you still have to keep in mind dissipation and other limitations for the BJT.)
Hopefully, that helps.
