Let us derive a formula for $a(n)$.
We will employ the inclusion-exclusion principle to get $a(n)$ as the number of partitions not satisfying any of the properties $P_i =$ "elements $i$ and $i+1$ get into the same part", $i\in[n-1]$.
Any collection of properties describe some maximal chains of consecutive elements going into the same parts. For example, $P_1$, $P_2$, and $P_4$ together describe chains $(1,2,3)$ and $(4,5)$, from which the elements should go into the same parts to satisfy those properties.
It is convenient to describe those maximal chains as a nonnegative composition $(k_0,l_1,k_1,\dots,l_c,k_c)$ of $n$, where the numbers $k_0\geq 0, l_1\geq2, k_1\geq0, \dots, l_c\geq2, k_c\geq0$ are the lengths of the consecutive chains forming $1,2,\dots,n$ such that those of lengths $l_1, \dots, l_c$ are resulted from the properties. In the above example with $P_1$, $P_2$, and $P_4$ and, say, $n=6$ we have the composition $(0,3,0,2,1)$.
Clearly, $s:=(k_0,l_1,k_1,\dots,l_c,k_c)$ corresponds to a collection of $l_1-1 + \dots +l_c-1 = L-c$ properties, where $L:=l_1+\dots+l_c$. Also, the number of partitions of subsets of $[n]$ satisfying $s$ equals
$$\sum_{m=L}^n \binom{n-L}{m-L} B_{m-L+c},$$
where $m$ corresponds to the subset sizes and $B_{m-L+c}$ is Bell number.
Hence, by inclusion-exclusion we have
\begin{split}
a(n) &= \sum_{L=0}^n \sum_{c=0}^{[L/2]} \binom{n-L+c}{c} \binom{L-c-1}{L-2c} (-1)^{L-c} \sum_{m=L}^n \binom{n-L}{m-L} B_{m-L+c} \\
&=\sum_{k\geq 0} B_k \sum_{c\geq 0} \sum_{L\geq 2c} \binom{n-L+c}{c} \binom{L-c-1}{L-2c} (-1)^{L-c} \binom{n-L}{k-c} \\
&=\sum_{k\geq 0} B_k \sum_{c\geq 0} \binom{k}c \sum_{L\geq 2c} \binom{n-L+c}{k} \binom{L-c-1}{L-2c} (-1)^{L-c} \\
&=\sum_{k\geq 0} B_k \sum_{c\geq 0} \binom{k}c (-1)^{n+k} [x^{n-k}] x^c (1+x)^{-k-1} (1-x)^{-c} \\
&=\sum_{k\geq 0} B_k (-1)^{n+k} [x^{n-k}] (1+x)^{-k-1} \big(1+\frac{x}{1-x})^k \\
&=[x^n] \frac1{1-x} \sum_{k\geq 0} B_k \big(\frac{x}{1-x^2}\big)^k,
\end{split}
which matches the expression for $B(x)$ I gave in the comments.
