$\newcommand\R{\mathbb R}$Most likely, the integral 
$$I:=\int_0^t dt_1 \int_0^t dt_2 \int_0^t dt_3\, B(t-t_3)F(t_3-t_2)
S(t-t_2)F(t_2-t_1)F(t_1)$$
cannot be expressed in terms of products and convolutions -- because the finite interval $[0,t]$ is not a group (or even a semigroup), and the restrictions $t_2\in[0,t]$ and $t_3\in[0,t]$ are unjustified, since the integrand contains no factors depending only on $t_2$ or only on $t_3$. Since the functions $B,F,S$ are zero on $(-\infty,0)$, the correct(ed) form of the integral $I$ seems to be 
\begin{align*}
	\tilde I&:=\int_\R dt_1 \int_\R dt_2 \int_\R dt_3\, B(t-t_3)F(t_3-t_2)
S(t-t_2)F(t_2-t_1)F(t_1) \\
&=\int_\R {dt_2}\, (B*F)(t-t_2)S(t-t_2)(F*F)(t_2) \\ 
&=\Big(\big((B*F)S\big)*(F*F)\Big)(t),
\end{align*}
so that $\tilde I$ is expressed in terms of products and convolutions. 

---

The similarly corrected integral added later to the OP, 
\begin{align*}
	J&:=\int_\R dt_1 \int_\R dt_2 \int_\R dt_3\,  A(t-t_3)F(t_3-t_2)S(t-t_2)F(t_2-t_1)S(t_3-t_1)F(t_1),
\end{align*}
cannot be expressed in terms of products and convolutions. 

The reason is the following *shift* observation, **which addresses a comment by LSpice**: 

>If an integral like $\tilde I$ or $J$ can be expressed in terms of products and convolutions, then the horizontal shift of each of the involved functions by a real number $u$ results in the horizontal shift of the resulting function by $ku$, where $k$ is the number of convolutions in the representation of the resulting function. 

E.g., $k=3$ in the above representation of the integral $\tilde I$. 

Let now $j(u,t)$ denote the integral $J$ with $A,F,S$ each being the density function of the normal distribution with mean $u$ and variance $1$. Then for all real $u$ and $t$ 
\begin{equation}
	j(u,t)=\frac1{8 \sqrt{2}\, \pi ^{3/2}}\, e^{-(t^2+6 t u+10 u^2)/4}
\end{equation}
and hence 
\begin{equation}
	\frac{j(u,t)}{j(0,t-k u)}=e^{u ((k^2-10) u-2 (k-3) t)/4},
\end{equation}
which is not identically $1$ in $(u,t)$, for any given integer $k$. So indeed, the integral $J$ cannot be expressed in terms of products and convolutions.