$$\sum^{\infty}_{n=1}\frac{a[n]}{n!}=\frac{a[1]}{1}+ \frac{a[2]}{2}+ \frac{a[3]}{6}+ \frac{a[4]}{24}+\dots=I$$
$$\sum^{\infty}_{n=1}\frac{a[n]}{(n-1)!}=\frac{a[1]}{1}+ \frac{a[2]}{1}+ \frac{a[3]}{2}+ \frac{a[4]}{6}+\dots=J$$
Is there a way to express $J$ only in terms of $I$?
$a[n]$ are polynomials in terms of X that satisfy the following relation: $$ \frac{d}{d X} a[n] = n (n-1) a[n-1] - n \frac{d}{d X} a[n-1] $$
One possible approach if $a_n$ ever becomes known in some sense would be through exponential generating functions. Define $$ f_I(x) = \sum_{n=1}^\infty \frac{a_n}{n!} x^n $$ and note that $$ \begin{split} f_i'(x) &= \frac{d}{dx} \left[ \sum_{n=1}^\infty \frac{a_n}{n!} x^n \right] = \sum_{n=1}^\infty \frac{d}{dx} \left[ \frac{a_n}{n!} x^n \right] \\ &= \sum_{n=1}^\infty \frac{a_n}{(n-1)!} x^{n-1} = \frac{1}{x} \sum_{n=1}^\infty \frac{a_n}{(n-1)!} x^n \\ &= \frac{f_J(x)}{x} \end{split} $$ and knowing the relationship between generating functions $f_J(x) = xf_I'(x)$ will help compute the underlying sequences as well.
That said, analytically computing one sequence in terms of the other is not possible without further information on $a_n$...