One must surely realize that adding a current source does not make much of a circuit given the infinite source resistance providing isolation. However, a voltage source has zero source resistance - there is no isolation at all. A voltage source is not part of the device. It is external to the device, and its addition makes ...what? And who says the device has the same characteristics as the circuit to which it is embedded? My claim is that good transistor circuit design makes the value of Vbe not relevant.
It is also true that at the same temperature, the base current, Vbe, and collector current of a BJT will be the same as it would be if the transistor were isolated from the circuit and run only on current sources providing the same current as provided by the voltage sources. How could it be any different? The relationship between base current and collector current will still be the same. The current being forced to flow in the junction generates the same voltage as before, and that cannot be explained as a voltage divider since voltage division in general does not work when a resistance is infinite. Also, the most recognizable and useful (since it provides the basis for constructing the load line) is the static characteristic where the base is driven by a stepped current source that clearly shows the relationship between between base current and collector current. You might say this really only shows the operation of the measurement circuit; however, it may also reflect the manufacturers' desire that the device will actually be used in a real circuit.
The terminal characteristics of a device are sufficient information to make the device useful in circuit design. For large-signal analysis the only useful model is the transistor as a current amplifier. If as you say above, the addition of a current source to control the base current makes a current amplifying circuit, then that is fine because large signal circuits drive the base with a current anyway. For small-signal high-frequency analysis the most useful model is the transistor as a transconductance amplifier because the math just works out to be simpler that way.
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