I suggest to show this inductively via showing that the difference $a(n+1)-a(n)$ always equals $b(n+1)-b(n)$.
As mentioned in the comments, the set $S(n)$ of $(4,3,1)$-avoiding partitions of $n$ is exactly the set of those for which the number of distinct parts is at most $3$, and if it equals $3$, then the smallest and second smallest parts should differ by $1$. Define now a map on partitions by increasing exactly one instance of the smallest part by $1$. This gives a "nearly injective" map from $S(n)$ into $S(n+1)$; indeed, the only "double hits" are partitions of the form $[a^k, (a-1)^m]$ for $a\ge 3$ and $k,m\ge 1$ (here $a^k$ should denote $k$-fold occurrence of $a$), since those are the images of both $[a^{k-1}, (a-1)^{m+1}]$ and $[a^k, (a-1)^{m-1}, a-2]$. Denote the set of these double hit partitions by $A$ and note that via conjugating Ferrer diagrams, $A$ is in bijection with the set of partitions of $n+1$ of the form $[a,b^k]$ with $a<b$ and $k\ge 2$, i.e., with the set of solutions to $n+1=a+kb$ with $1\le a<b$ and $k\ge 2$. For each $b\in \{1,\dots, \lfloor n/2\rfloor\}$, there is exactly one such solution, except for the divisors $b$ of $n+1$ (for which there is none). I.e., $$|A| = \lfloor n/2 \rfloor - d(n+1),$$ with $d$ the divisor function.
Moreover, the set $B$ of partitions in $S(n+1)$ and not in the image are exactly the trivial partition $[1,\dots, 1]$ and the partitions $[a,\dots, a, 1\dots, 1]$ with $a\ge 3$ (and at least one instance of both $a$ and $1$). From this, it's obvious that $$|B|= 1+\sum_{a=3}^n \lfloor \frac{n}{a}\rfloor,$$ since indeed $\lfloor \frac{n}{a}\rfloor$ is just the number of possible multiplicities of the given value $a$ in a partition of the form above.
Thus, the difference $a_{n+1}-a_n$ is exactly $|B|-|A|$.
On the other hand, the difference $b(n+1)-b(n)$ is easily calculated (if you want, distinguish at first between even and odd $n$) as $$b(n+1)-b(n) = 1+\sum_{i=1}^{\lfloor n/2 \rfloor} \lfloor \frac{n-i}{i+1}\rfloor = 1+(\sum_{i=1}^{\lfloor n/2 \rfloor} \lfloor \frac{n+1}{i+1}\rfloor ) - \lfloor n/2\rfloor = 1+(\sum_{i= 1}^{\lfloor n/2 \rfloor} \lfloor \frac{n}{i+1}\rfloor ) - |A| - 1,$$ the last equality coming from the fact that $\lfloor \frac{n}{i+1}\rfloor$ and $\lfloor \frac{n+1}{i+1}\rfloor$ differ iff $i+1\ge 2$ is a divisor of $n+1$ and now I think it's becoming evident that this is the same as the difference |B|-|A| calculated above (maybe up to some small index mistake due to parity of $n$; I'll check this later).