Earliest Arrival Time Let $\tau_m = \min(\tau_1,\tau_2)$ where $\tau_i,i=1,2$ are i.i.d. random failure times, with survival function $\Pr(\tau>t)=g(t)$ for some well-behaved continuous function $g$ defined on $\mathbb{R}^+$. Is there any relationship between $ \mathbb{E}[\tau_m] $ and moments of $\tau$ ? 
 A: Below I will follow slightly different notation which aligns with standard definitions in probability and set the cumulative distribution function $F(t) = \mathbf P( \tau_1 \leq t)$, so that in your notation $F(t) = 1-g(t)$, and then the probability density function is $f(t) = \frac{d}{dt}F(t)$.
Similarly I will denote $F_m, f_m$ for the CDF and PDF of the minimum $\tau_m = \min(\tau_1,\tau_2)$.
From independence we have
\begin{align*}
\mathbf F_m(t) & = 1 - \mathbf P(\tau_m > t) \\& = 1-\mathbf P( \tau_1 > t, \, \tau_2 > t) \\
& = 1- \mathbf P( \tau_1 > t) \mathbf P( \tau_2 > t) \\
& = 1 -g(t)^2 \\
& = 1 - \big(1 - F(t)\big)^2 \\
& = 2F(t) - F(t)^2.
\end{align*}
Now we note that the expectation is defined as
$$ \mathbf E[\tau_m] = \int t f_m(t) dt = \int t \left( \frac{d}{dt}F_m\right)(t) dt$$
So in particular
\begin{align*}
\mathbf E[\tau_m] & = \int t \frac{d}{dt}\left( 2F(t) - F(t)^2 \right) \,dt \\
& = \int t \big( 2f(t) - 2f(t)F(t) \big) dt \\
& = 2 \int t f(t) dt - 2 \int t\, f(t) F(t)dt \\
& = 2 \mathbf E[\tau]  - 2\int t\,  f(t)F(t) dt .
\end{align*}
Unfortunately from there I do not think you can simplify the first integral without specific knowledge of the distribution.
Example
Consider the case where $\tau_1,\tau_2 \sim \text{Unif}[0,1]$. In particular we have
$$F(t) = t, \qquad f(t) = 1, \qquad 0 \leq t \leq 1.$$
The above calculation gives
$$\mathbf E[\tau_m] = 1 - 2\int_0^1 t^2dt = \frac13.$$
