Help with a $z$-transfrom Problem
You might rewrite your signal as

$$x(t)=(1-e^{-2t})u(t-1)=(1-e^{-2t})u(t)-(1-e^{0})\delta(t)=(1-e^{-2t})u(t)$$

The idea is to express the signal $x(t)$ as the difference of two signals. The first signal has the step function starting at $t=0$, and the second function removes the contribution of the step at the sample $t=0$. In your case, even the function has $x(0)=0$, so it does not matter, if you have the step function starting at $t=0$ or $t=1$. So, you can just apply your solution above and you have it.

To explain my steps in a bit more detail: First note, that 
$$u(t)=u(t-1)+\delta(t).$$(we are in a discrete setting with $t\in\mathbb{Z}$). Therefore $$u(t-1)=u(t)-\delta(t).$$ Let's insert this into your signal (we take a general $h(t)$. In your case you'd have $h(t)=1-\exp(-2t)$:

$$x(t)=h(t)u(t-1)=h(t)(u(t)-\delta(t))=h(t)u(t)-h(t)\delta(t)$$

The last part, i.e. $h(t)\delta(t)$ becomes equal to $h(0)\delta(t)$ due to the sifting property of the Dirac. So, in general you have 

$$\mathcal{Z}\{h(t)u(t-1)\}=\mathcal{Z}\{h(t)u(t)\}-\mathcal{Z}\{h(0)\delta(t)\}=H(z)-h(0),$$

where $H(z)$ is the Z-transform of $h(t)$, and we assume that $h(t)$ is causal, i.e. $h(t)=h(t)u(t)$. (This property is implicitely assumed in the standard Z-transform).

In your case, you have $h(0)=0$, so the second term vanishes.