This is an answer to the request in some comments to show how to arrive at the formula for the entries of $L^h$
Here is an example how we can find the entries of $L^h$ symbolically. I do with the $4 \times 4 $ triangular matrix $L$
$$ M(h) = \exp( h \cdot \log (L)) = L^h $$
and get for $M$
$$ M(h) = \left[ \begin{array} {rrrr}
1 & . & . & . \\
1h & 1 & . & . \\
\frac 12( 1 h^2+ 1 h) & 1h & 1 & . \\
\frac 16( 1h^3+ 3 h^2+ 2 h) & \frac 12 (1h^2+ 1h) & 1h & 1
\end{array} \right]$$
Here a trained eye recognizes the Stirling numbers 1st kind as coefficients at the powers of $h$ and because the structure of the matrix has this constant diagonals it is easy to make the formula for the transfer:
$$ M(h)\cdot S_k = S_{k+h} $$
One more step shows, that the evaluation of the polynomials in the entries leads to binomial numbers, which is a well known property of the Stirling numbers first kind (the vandermonde-matrix LDU-decomposes into the matrices of Stirling-numbers 2st kind and of binomial coefficients and thus reduces by the multiplication with the matrix of Stirling numbers 1'st kind (which is the inverse of the 2nd-kind matrix) to binomials)
I had fun to proceed a bit. Factorizing the smbolic entries, assuming the hypothese that we have always the Stirling numbers 1st kind and the fractional cofactors the reciprocal fatorials give
$$ M(h) = \left[ \begin{array} {llll}
1 & . & . & . \\
1(h) & 1 & . & . \\
\frac 12( h(h+1)) & 1(h) & 1 & . \\
\frac 16( h(h+1)(h+2)) & \frac 12 ( h(h+1)) & 1(h) & 1
\end{array} \right]$$
and this gives immediately the binomial coefficients
$$ M(h) = \left[ \begin{array} {cccc}
1 & . & . & . \\
\binom{h}{1} & 1 & . & . \\
\binom{h+1}{2} & \binom{h}{1} & 1 & . \\
\binom{h+2}{3} & \binom{h+1}{2} & \binom{h}{1} & 1
\end{array} \right]$$
and a routine could solve the problem given the vector S of dimension n in the following way:
T = 0*S ; \\ initialize an empty array of size of S
b = 1; \\ contains the current binomial
for(j=1,n,
for(k=j,n, T[k]+=S[k+1-j]*b);
b *= (h+j-1)/j;
);
return(T);
So we have $n^2/2$ operations by the looping.