I edit my post to answer Carlo Beenakker's remark and also because I would like to add a reference, possibly more accurate than the two below. Theorem 7.1 p.13 of

A. Adelberg, A finite difference approach to degenerate Bernoulli and Stirling polynomials, Discrete Math. 140 (1995), 1-21,

states that, for $s$ a complex number,
$$\left(\frac{(1+t)^{y}-1}{ty}\right)^{s}=\sum_{n=0}^{\infty}B_{n,s}(y)t^{n},$$
where $B_{n,s}(y)$ is a specific polynomial defined from divided differences of binomial coefficients. Letting $y$ tend to 0 gives
$$(\log(1+t))^{s}=t^{s}\sum_{n=0}^{\infty}B_{n,s}(0)t^{n},$$
where it can be check that, with $\psi_{n}$ the Stirling polynomial,
$$B_{n,s}(0)=\frac{(-1)^{n}}{n!}\frac{s}{s+n}\psi_{n}(s+n-1).$$
Hence, the seeked expansion at $t=0$ for a real power of $\log(1+t)$ is given by
$$(\log(1+t))^{s}=t^{s}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\frac{s}{s+n}\psi_{n}(s+n-1)t^{n}.$$

The formula, written in a different form and with a complex exponent, appears on p.642 of this paper where generalized Stirling functions are used. These generalized Stirling functions are also studied in

Generalized Stirling Functions of Second Kind and Representations of Fractional Order Differences via Derivatives, P.L. BUTZER, A.A. KILBAS, J. TRUJILLO, Journal of Difference Equations and Applications, 2003 Vol. 9 (5), pp. 503–533

see in particular formula (3.9) there, to be compared with the definition of the Stirling polynomials. The link between the Stirling polynomials in your question and the generalized Stirling functions in these references shouldn't be too complicate to derive.