I'm wondering how to go about finding the equation that defines this locus. In short, the locus is all the points the same distance from the closest point on $e^x - 1$ as $(0,1)$. The setup in GeoGebra also lets me change the function it depends on easily which produces very interesting results for almost every function (for a straight line, it produces a parabola).
The distance from any point $P(x, y)$ on the required locus to the fixed point $Q(0, 1)$ is given by $$PQ^2=x^2+(y-1)^2$$ The distance from the same point $P(x, y)$ to a point $T$ on the given curve $(t, e^t-1)$ is given by $$s=PT^2=(x-t)^2+z^2$$ where $z=y-e^t+1$ hence $y-1=z+e^t-2$.
The point $T$ is closest to $P$ when $s$ is a minimum wrt $t$ : $$\frac{ds}{dt}=2(x-t)(-1)+2z(-e^t)=0$$ $$x=t-ze^t$$
The required locus is defined by $PQ^2=PT^2$ : $$x^2+(y-1)^2=(x-t)^2+z^2$$
Substitute to eliminate $x, y$ leaving an equation in $z, t$ : $$(t-ze^t)^2+(z+e^t-2)^2=z^2e^{2t}+z^2$$ $$t^2-2tze^t+e^{2t}+4-2(ze^t-2z-2e^t)=0$$ $$z=\frac{t^2+(e^t-2)^2}{2[(t-1)e^t+2]}$$
Co-ordinates of points on the locus can be obtained in terms of parameters $t$ and $z(t)$ :
$$x=t-ze^t, y=z+e^t-1$$