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)
$=\dfrac{\pi t}{2}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\pi t^{2n+2k+3}}{4^{n+k+2}(n+k+1)!\left(\dfrac{3}{2}\right)_{n+k+1}(k+1)}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\pi t^{2n+2k+1}}{4^{n+k+1}(n+k)!\left(\dfrac{3}{2}\right)_{n+k}\left(k+\dfrac{1}{2}\right)}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
$=\dfrac{\pi t}{2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\pi t^{2n+2k+3}(1)_k}{2^{2n+2k+3}3(2)_{n+k}\left(\dfrac{5}{2}\right)_{n+k}(2)_k}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\pi t^{2n+2k+1}\left(\dfrac{1}{2}\right)_k}{2^{2n+2k+1}(1)_{n+k}\left(\dfrac{3}{2}\right)_{n+k}\left(\dfrac{3}{2}\right)_k}-\dfrac{\pi(\gamma+\log t)\sinh t}{2}$
\begin{align*}\mathcal{L}\{f\}({s})\cdot\mathcal{L}\{g\}({s})&=\left(\int_0^\infty e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^\infty e^{-{s}u}{g}(u)\,du\right)\\
&=\lim_{{{L}}\to\infty}\left(\int_0^{{L}}e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^{{L}}e^{-{s}u}{g}(u)\,du\right)
\end{align*}
By Fubini's theorem,
\begin{align*}\left(\int_0^{{L}}e^{-{s}t}{f}(t)\,dt\right)\left(\int_0^{{L}}e^{-{s}u}{g}(u)\,du\right)&=\int_0^{{L}}\int_0^{{L}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du\\
&=\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du,
\end{align*}
where $R_{{L}}$ is the square region $$0\leq t\leq {{L}},\qquad 0\leq u\leq {{L}}.$$
Let $T_{{L}}$ be the triangular region
$$0\leq t\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad t+u\leq {{L}}.$$
Provided that ${f}$ and ${g}$ are bounded by exponential functions,
$$\lim_{{{L}}\to\infty}\iint_{R_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\lim_{{{L}}\to\infty}\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du.$$
Now, the function
$$\varphi(v,u)=(v-u,u)$$
maps $D_{{L}}$ bijectively onto $T_{{L}}$, where $D_{{L}}$ is the triangular region
$$0\leq v\leq {{L}},\qquad 0\leq u\leq {{L}},\qquad v\geq u.$$
The component functions of $\varphi$ are
$$t(v,u)=v-u\qquad\mbox{and}\qquad u(v,u)=u,$$
so the Jacobian of $\varphi$ is
$$J\varphi=\det\begin{bmatrix}t_v&t_u\\u_v&u_u\end{bmatrix}=\det\begin{bmatrix}1&-1\\0&1\end{bmatrix}=1.$$
Hence,
$$\iint_{T_{{L}}}e^{-{s}(t+u)}{f}(t){g}(u)\,dt\,du=\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du.$$
By Fubini's theorem,
\begin{align*}&\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du=\int_0^{{L}}\int_0^ve^{-{s}v}{f}(v-u){g}(u)\,du\,dv\\
\\
&=\int_0^{{L}}e^{-{s}v}\int_0^v{f}(v-u){g}(u)\,du\,dv=\int_0^{{L}}e^{-{s}v}({f}\ast{g})(v)\,dv.
\end{align*}
Hence,
\begin{align*}\lim_{{{L}}\to\infty}\iint_{D_{{L}}}e^{-{s}v}{f}(v-u){g}(u)\,dv\,du&=\lim_{{{L}}\to\infty}\int_0^{{L}}e^{-{s}v}({f}\ast{g})(v)\,dv\\
\\
&=\int_0^\infty e^{-{s}v}({f}\ast{g})(v)\,dv\\
\\
&=\mathcal{L}\{{f}\ast{g}\}({s}).
\end{align*}
Best Answer
No, not because the convolution theorem doesn't hold, but because the convolution of $F,G$ isn't defined like you did. The correct one is $$F \ast G(s) = \frac{1}{2i\pi}\int_{r-i\infty}^{r+i\infty} F(z) G(s-z) dz$$ for any $r > 0, \Re(s) > r$ which is well-defined for example when assuming $f,g$ are bounded so their Laplace transform is $L^2$ on vertical lines.
Assuming $f,g$ are $C^1$ and they vanish at $t=0$ then $F,G$ are $L^1$ on vertical lines and for $c > r$ using absolute convergence / Fubini to change the order of integration $$\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty} F \ast G(s)e^{st}ds = \frac{1}{2i\pi} \int_{c-i\infty}^{c+i\infty}\frac{1}{2i\pi}\int_{r-i\infty}^{r+i\infty} F(s-z) G(z)e^{st}dz ds $$
$$= \frac{1}{2i\pi} \int_{c-i\infty}^{c+i\infty}\frac{1}{2i\pi}\int_{r-i\infty}^{r+i\infty} F(s-z) e^{(s-z)t}G(z)e^{zt}dz ds $$
$$= \frac{1}{2i\pi} \int_{r-i\infty}^{r+i\infty}\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty} F(s-z) e^{(s-z)t}G(z)e^{zt}ds dz$$
$$= \frac{1}{2i\pi} \int_{r-i\infty}^{r+i\infty}\frac{1}{2i\pi}\int_{c-r-i\infty}^{c-r+i\infty} F(u) e^{ut}G(z)e^{zt}du dz$$
$$= (\frac{1}{2i\pi} \int_{r-i\infty}^{r+i\infty} F(u) e^{ut}du )(\frac{1}{2i\pi}\int_{c-r-i\infty}^{c-r+i\infty}G(z)e^{zt} dz)$$ $$ = f(t)g(t)$$