*The following is a rephrasing of and an expansion on Joris Bierkens' answer. It serves as my note for my own benefit. It would be a bonus if it helps any other people.*

Define stochastic processes $\displaystyle y_t := \exp\Big(-\int_0^t r_s \,ds \Big)$. The subscript denotes the time variable dependency. Write $f(t) := \mathbb E [y_t]$. We apply the Ito's lemma recursively. In fact, the same procedure leads to the Ito version of the Taylor expansion to an arbitrary order.

Apply Ito's Lemma twice
\begin{align}
-(y_t-y_0) &=\int_0^tr_sy_sds \\
&=\int_0^tds\Big(r_0y_0+\int_0^s(y_udr_u+r_udy_u)\Big) \\
&=r_0t+\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u)+\int_0^tdB_u\,y_u\sigma_u(t-u)
\end{align}
since $dr_tdy_t=-r_ty_tdr_tdt=0$. The last integral comes from exchanging the order of two integrations.

Apply Ito's lemma again
$$d\big(y(\mu-r^2)\big) = y\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial t}\Big)dt+y\sigma\Big(\frac{\partial\mu}{\partial r}-2r\Big)dB.$$
Then interchange the order of integration,
\begin{align}
&2\int_0^tdu\ y_u(\mu_u-r_u^2)(t-u) \\
= & y_0(\mu_0-r_0^2)t^2+\int_0^tds\ y\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial s}\Big)(t-s)^2 + \int_0^t dB\,\sigma y\Big(\frac{\partial\mu}{\partial r}-2r\Big)(t-s)^2
\end{align}
where all the variables except $t$ in the integrand depends on $s$.

Take expectation of the above expression. As $\mathbb E \left[ y\Big(r^3-3\mu r+\mu\frac{\partial\mu}{\partial r}+\frac{\partial\mu}{\partial s}\Big)\right]$ is bounded,
$$f(t) = 1-r_0t+\frac12(\mu_0-r_0^2)t^2+O(t^3)=\exp\Big(-r_0t+\Big(\frac12\mu_0-r_0^2\Big)t^2+O(t^3)\Big)$$
as $t\searrow 0$.

It remains to fill in details regarding the interchange of integration and expectation by Fubini's theorem, which should be justified using a priori estimates on the stochastic quantities using the strong linear growth assumptions.