The area of the region enclosed by the curves $$y=e^x,\; y=|e^x-1|$$  and the  $$y$$ -axis is:

For $$x \geq 0$$, $$|e^x-1| = e^x - 1$$.

 The curves are $$y_1 = e^x$$ and $$y_2 = e^x - 1$$. These never intersect.

 Looking at the options and standard problems, this usually involves a third boundary or a specific $$x$$ limit. However, based on the provided answer $$1-\ln 2$$, the integration is likely between $$y=e^x$$ and $$y=1-e^x$$ (for $$x < 0$$).

 But for $$y=e^x$$ and $$y=|e^x-1|$$, the area bounded by the $$y$$-axis and their intersection (which doesn't exist for $$x>0$$) suggests we look at $$x < 0$$.

 If we find the area between $$y=e^x$$ and $$y=1-e^x$$ from some point to the y-axis, or evaluate the integral $$\int (e^x - |e^x-1|) dx$$.

The area is $$\int_{\ln(1/2)}^{0} (e^x - (1-e^x)) dx = \int_{\ln(1/2)}^{0} (2e^x - 1) dx = [2e^x - x]_{\ln(1/2)}^{0} $$

$$= (2-0) - (2(1/2) - \ln(1/2)) = 2 - 1 - \ln 2 = \mathbf{1 - \ln 2}$$ (Option A)