Unfortunately, the convolution in above cannot directly solve as it contains some divergent integrals, so you should consider on this approach instead.
With the result of http://eqworld.ipmnet.ru/en/auxiliary/inttrans/LapInv7.pdf,
$\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\dfrac{\log s}{s^2-1}\right\}$
$=\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\dfrac{\log s}{s^2\left(1-\dfrac{1}{s^2}\right)}\right\}$
$=\mathcal{L}^{-1}_{s\to t}\left\{\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{\log s}{s^{2n+2}}\right\}$
$=\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=1}^{2n+1}\dfrac{t^{2n+1}}{(2n+1)!k}-\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{t^{2n+1}(\gamma+\log t)}{(2n+1)!}$
$=\dfrac{\pi t}{2}+\dfrac{\pi}{2}\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{t^{2n+1}}{(2n+1)!2k}+\dfrac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{t^{2n+1}}{(2n+1)!(2k+1)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
$=\dfrac{\pi t}{2}+\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{\pi t^{2n+1}}{4^{n+1}n!\left(\dfrac{3}{2}\right)_nk}+\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{\pi t^{2n+1}}{4^{n+1}n!\left(\dfrac{3}{2}\right)_n\left(k+\dfrac{1}{2}\right)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$ (according to http://mathworld.wolfram.com/PochhammerSymbol.html)
$=\dfrac{\pi t}{2}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\pi t^{2n+2k+3}}{4^{n+k+2}(n+k+1)!\left(\dfrac{3}{2}\right)_{n+k+1}(k+1)}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\pi t^{2n+2k+1}}{4^{n+k+1}(n+k)!\left(\dfrac{3}{2}\right)_{n+k}\left(k+\dfrac{1}{2}\right)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
$=\dfrac{\pi t}{2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\pi t^{2n+2k+3}(1)_k}{2^{2n+2k+3}3(2)_{n+k}\left(\dfrac{5}{2}\right)_{n+k}(2)_k}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\pi t^{2n+2k+1}\left(\dfrac{1}{2}\right)_k}{2^{2n+2k+1}(1)_{n+k}\left(\dfrac{3}{2}\right)_{n+k}\left(\dfrac{3}{2}\right)_k}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
The Gamma function has poles at $z=0$ and at the negative integers. When $z=-k$, then the residue at that pole is $(-1)^k/k!$. Now, the inverse transform is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \Gamma(p+1+T s) e^{s (t-T \log{N})} $$
The poles of this Gamma function are at $p+1+T s = -k$, or $s=-(p+1+k)/T$ for $k \in \{0,-1,-2,\ldots \}$. Thus, the ILT is
$$e^{-(p+1)(t-T \log{N}))/T} H(t-T \log{N})\sum_{k=0}^{\infty} \frac{(-1)^k}{k!} e^{-k (t-T \log{N})/T} $$
where $H$ is the Heaviside function, or, summing the series,
$$e^{-(p+1)(t-T \log{N}))/T} \exp{\left [-e^{-(t-T \log{N})/N}\right ]} H(t-T \log{N}) $$
which, aside from the Heaviside factor, agrees with your expected result.
Best Answer
Write
$$ \frac{a\sinh\left[\frac pc (L - x)\right]}{p\sinh\left[ \frac {pL}c\right]} = \frac ap \left\{ \exp\left[ \frac {pL}c\right] - \exp\left[ -\frac {2pL}c + \frac{px}{c}\right]\right\} \left\{ 1 - \exp \left[ -\frac{2pL}{c}\right]\right\}^{-1}. $$
Use the binomial theorem to expand the factor raised to the power of $-1$ (not worrying about whether the resulting series converges or not... we'll see later that we get a sensible answer):
$$ \begin{align} \frac{a\sinh\left[\frac pc (L - x)\right]}{p\sinh\left[ \frac {pL}c\right]} &= \frac ap \left\{ \exp\left[ \frac {pL}c\right] - \exp\left[ -\frac {2pL}c + \frac{px}{c}\right]\right\} \sum_{k=0}^\infty \exp \left[ -\frac{2pLk}{c}\right] \\ &= a\sum_{k=0}^\infty \left\{ \frac 1 p \exp\left[ -p \left( \frac xc + \frac{2Lk}{c}\right) \right] - \frac 1 p \exp\left[ -p \left( -\frac x c + \frac{2L(k+1)}{c}\right)\right]\right\}. \end{align} $$ It's not hard to check that, for $\lambda > 0$, $$ \mathcal L_{t \to p} (H(t - \lambda)) = \frac{e^{-\lambda p}}{p}, $$ where $H$ is the Heaviside function. So assuming that $0 < x < L$, $$ \frac{a\sinh\left[\frac pc (L - x)\right]}{p\sinh\left[ \frac {pL}c\right]} = \mathcal L_{t \to p} \left\{ a\sum_{k=0}^\infty \left[ H\left( t - \frac{x + 2Lk}{c} \right) - H\left( t - \frac{-x + 2L(k+1)}{c} \right)\right]\right\}. $$ So the inverse Laplace transform is $$ a\sum_{k=0}^\infty \left[ H\left( t - \frac{x + 2Lk}{c} \right) - H\left( t - \frac{-x + 2L(k+1)}{c} \right)\right]. $$ The $k$th term of this series is $1$ if $t \in \left(\frac{x + 2Lk}{c}, \frac{-x + 2L(k + 1)}{c} \right)$ and $0$ otherwise. You can think of each term as initially starting in the "off" position, turning "on" for a finite amount of time, then turning "off" again. This series also converges for all $x \in (0, L)$ and $t > 0$ since there are only finitely many terms that are "on" at any such $x$ and $t$.
Please let me know if you spot any mistakes.