$\newcommand\R{\mathbb R}$
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*}
probably cannot be expressed in terms of products and convolutions. 

A way to check it may be by the following *shift* observation, **which addresses a comment by LSpice**: 

>Suppose we want to show that a function $f$ cannot be expressed in terms of products and convolutions of functions $f_1,\dots,f_n$. To do this, we build all possible binary trees, as was apparently suggested by Terry Tao. The nodes of such a tree are nonempty subsets of the set $[n]=\{1,\dots,n\}$. To each node $J$ there corresponds an expression in terms of products and/or convolutions of the functions $(f_j\colon j\in J)$. 
The root of the tree is the set $[n]$. Each node is split into two branches, whose roots are nonempty subsets, which become the nodes next up the tree. Each node is is marked either by $p$ (for "product") or $c$ (for "convolution"), depending on whether the daughter expressions of the node get multiplied or convolved. 

>Then the functions $f_1,\dots,f_n$ get shifted horizontally, starting with the leaves of the tree. If two adjacent leaves are the branches of a $p$-node, their corresponding functions get shifted by the same real number, say $u$. After that, the two leaves get deleted and their parent node is marked by the same shift $u$. If two adjacent leaves are the branches of a $c$-node, their corresponding functions get shifted by two arbitrary real numbers, say $u_1$ and $u_2$. After that, the two leaves get deleted and their parent node is marked by shift $u_1+u_2$. This process then continues down to the root. For every $p$-node encountered during this process, the shifts assigned to the daughter branches get equalized, say by imposing the corresponding equality constraint. Finally, at the root of the tree, we check if the function $f$ has the resulting shift property. 

As a comparatively simple case, suppose that a function $f$ can be expressed in terms of products and convolutions of functions $f_1,\dots,f_n$ as follows: 
\begin{equation}
	f=\Big(\prod_{j\in J_1}f_j\Big)*\cdots*\Big(\prod_{j\in J_m}f_j\Big), \tag{1}
\end{equation}
where $\{J_1,\dots,J_m\}$ is a partition of the set $\{1,\dots,n\}$. 
Suppose next that, for each $k=1,\dots,m$ and each $j\in J_k$, the function $f_j$ is shifted horizontally by a real number $u_k$. Then $f$ gets shifted by $u_1+\dots+u_k$. 

For any given functions $f_1,\dots,f_n$, there may be only finitely many representations of $f$ in the form (1); so, we can try and exhaust all of them.  

Using this test, it is easy to see that, e.g., the integral $J$ cannot be expressed as the pure convolution of the functions $(f_1,\dots,f_6):=(A,F,S,F,S,F)$ (which corresponds to the partition of the set $\{1,\dots,6\}$ into the singleton sets).  
Take here, for instance, each of the functions $A,F,S$ to be the density function of the normal distribution with mean $u$ and variance $1$. Other combinations of the form (1) can be considered similarly. Since here we have $f_2=f_4=f_6$, I have counted 10 more distinquishable partitions in this case.