t-axis translation rule can be proven easily using definition of laplace transform: $$\mathcal{L}{\{u(t-a)f(t-a)\}} = e^{-as}F(s)$$
Does this mean the inverse transform of $e^{as}F(s)$ is $u(t+a)f(t+a)$ ?
When I took laplace transform of $u(t+a)f(t+a)$, it is not simplifying...
so what is the inverse laplace transform of $e^{as}F(s)$ ?
My work:
$$\mathcal{L}{\{u(t+a)f(t+a)\}} = \int\limits_0^{\infty}e^{-st}u(t+a)f(t+a)\,dt $$
Letting $t+a=t_1$:
$$e^{as}\int\limits_a^{\infty}e^{-st_1}u(t_1)f(t_1)\,dt_1 $$
Not really sure of the next step. Any help?
($u(t_1)$ doesn't kill the graph between $0$ and $a$, so I can't let the bottom limit be $0$)
This is no complete answer, but there are some indications that your proposed identity does not hold. I think it's useful to have a look at the proof for the first identity you're stating. (Throughout my answer, $a>0$.) \begin{align} \mathcal{L}\left\{u(t-a)f(t-a)\right\} &=\int_0^{\infty}u(t-a)f(t-a)e^{-st}dt\\ &=\int_a^{\infty}f(t-a)e^{-st}dt\\ &=e^{-as}\int_0^{\infty}f(\tau)e^{-s\tau}dt\\ &=e^{-as}F(s) \end{align} where I used $\tau=t-a$. To go to the second line, I used that $u(t-a)=$ for $t<a$. Notice how this does not apply in your case, where, instead, you could've used that \begin{align} \int_0^{\infty}u(t+a)f(t+a)e^{-st}dt=\int_0^{\infty}f(t+a)e^{-st}dt. \end{align} You could also try to calculate the inverse Laplace transform directly, \begin{align} \mathcal{L}^{-1}\left\{e^{as}F(s)\right\} &=\int_0^t \delta(\tau+a)f(t-\tau) d\tau\\ \end{align} where I used $\mathcal{L}^{-1}\left\{G(s)F(s)\right\}=\int_0^tg(\tau)f(t-\tau)d\tau$ and $\mathcal{L}^{-1}\left\{e^{as}\right\}=\delta(t+a)$. The above integral is zero as the delta function peaks at $\tau=-a$, which is outside the integration range.