Still not a complete answer, but a method to be completed or improved.

Denote $f_n(t)=\frac{2 - e^{-t}\left ( 2 + 2t+t^2 \right )}{t^3}\left ( \frac{1-e^{-t}}{t} \right )^{n-1}$. Assume that we have found the coefficients $c_1,c_2,\dots,c_n$ and polynomials $g_1,\dots,g_n$ such that $$f_n(t)=2t^{-n-2}+\sum_{k=1}^n(g_k(t^{-1})e^{-kt})'+\sum_{k=1}^n c_k \frac{e^{-kt}}t.\,\,(*)$$
Then $\sum c_k=0$ (else $f_n$ would have a non-zero residue at 0, which is absurd). We have $\int_0^\infty \sum_{k=1}^n c_k \frac{e^{-kt}}t dt=-\sum c_k\log k$ by [Frullani integrals][1].
The integral of the other part of our sum $(*)$ is minus the limit at zero of the function $-\frac2{n+1}t^{-n-1}+\sum g_k(t^{-1})e^{-kt}$. The limit must exist, since the initial integral converges. For fixed $k$, we get an equation for $g_k$ and $c_k$: $-kg_k(y)-y^2g_k'(y)+c_ky=q_k(y)$ for certain explicit polynomial $q_k$ (I denote $y=1/t$ here). Note that this $q_k$ has not too many monomials, about 3 or 4. This equation has unique solution which may be found by equating coefficients. It appears to collect everything and get an answer. Note that for the answer we need only the constant term of the Laurent series $g_k(t^{-1})e^{-kt}$ and the value of $c_k$.

  [1]: http://mathworld.wolfram.com/FrullanisIntegral.html