Inverse Laplace transform shifting error

Question

I am doing the inverse Laplace transform of the function: $\frac{e^{-s}}{s-1}$.

I am solving and receiving the answer: $e^t\mathcal{U}(t-1)$, however the correct answer is $e^{t-1}\mathcal{U}(t-1).$

Im failing to see where the $-1$ in the exponential is coming from, and I have attempted several times.

I am using the equation $\mathcal{L}^{-1}[e^{-as}F(s)] = \mathcal{U}(t-a)f(t-a)$;

Answer 1

$F(s)={1\over s-1}\implies f(t)=e^t$ so using the theorem you want $u(t-a)\color{blue}{f(t-a)}=u(t-1)f(t-1)=u(t-1)e^{t-1}$

Answer 2

I prefer to use the definition then tables to solve the inverse Laplace transform. \begin{align} \mathcal{L}^{-1}\{e^{-s}/(s - 1)\} &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{-s}e^{st}}{s- 1}ds\\ &= \sum\text{Res}\\ &= \lim_{s\to 1}(s-1)\frac{e^{s(t-1)}}{s-1}\\ &= e^{t - 1}\mathcal{U}(t - 1) \end{align} Then from the definition, we see where the $t - 1$ comes from.

Just to note, I don't know why the unit step comes into play when we have a shift. I have asked a question on it here 10 days agos but no answer yet.