This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds
$$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
$$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$
where $O(1)$ denotes quantities that are absolutely bounded. First of all,
$$ f_n(t) = \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d[x] = 
\int_{0}^\infty \mu^x(1-\mu^xt)^n\ dx - \int_{0-}^\infty \mu^x(1-\mu^xt)^n\ d\langle x\rangle,$$
where $x=[x]+\langle x\rangle$ is the decomposition into integral and fractional parts. Evaluating the first integral on the right, and applying integration parts on the second integral, we obtain
$$ f_n(t) = \frac{1-(1-t)^{n+1}}{(n+1)(-t\log\mu)}+ (1-t)^n+\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx.$$
Here
$$ \frac{d}{dx}(\mu^x(1-\mu^xt)^n) = (\log\mu)\mu^x(1-\mu^xt)^{n-1}(1-(n+1)t\mu^x) $$
changes sign only once, namely where $\mu^x=\frac{1}{(n+1)t}$, hence the last integral can be estimated readily by $0\leq \langle x\rangle < 1$ as
$$\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx\ll (1-t)^n+\frac{1}{nt}.$$
It follows that
$$ (n+1)tf_n(t)=\frac{-1}{\log\mu}+O_\mu(n(1-t)^{n})+O(1), $$
which implies the claim above, noting that $\lim_{n\to\infty}n(1-t)^{n}=0$.

Based on the above argument I doubt that $\lim_{n\to\infty}n f_n(t)$ exists. More precisely, I don't think that
$$ \lim_{n\to\infty} n\int_0^\infty \frac{d}{dx}(\mu^x(1-\mu^xt)^n)\ \langle x\rangle\ dx $$
exists, because the integral is very sensitive on the fractional part of $\log_\mu\frac{1}{(n+1)t}$ where the derivative in the integrand changes its sign.

**EDIT:** It is easy to verify in an elementary fashion that $\lim_{n\to\infty} nf_n(t)$ does not always exist. Let $t:=\frac{1}{r^2}$ and $\mu:=\frac{1}{r^2}$, where $r>0$ is an integer. Then for any integer $\ell>0$ we have
$$n=r^{2\ell} \ \Longrightarrow\  nf_n(t) > n\mu^{\ell-1}(1-\mu^{\ell-1}t)^n=r^2(1-1/n)^n \gg r^2,$$
$$n=r^{2\ell+1}\  \Longrightarrow\  nf_n(t) \ll \max_{k=\ell,\ell-1}\ n\mu^k(1-\mu^kt)^n\ll r^3 2^{-r} + r \ll r.$$
Hence $\lim_{n\to\infty} nf_n(t)$ does not exist when $r$ is sufficiently large.