Let $\mu_t = (\varphi_t)_{\sharp} \mu$ denote the image of the measure $\mu$ by the flow of $a$ (where $\mu$ can be Lebesgue measure).
It is well-known that the family of measures $\{\mu_t\}_{t\in \mathbb R}$ satisfies Liouville equation (aka continuity equation)
$$
\partial_t \mu_t + \operatorname{div\,} (a \mu_t) = 0 \tag{1}
$$
in the sense of distributions. Indeed, for any smooth compactly supported function $f=f(t,x)$
$$
\iint (\partial_t f(t,y) + a(t,y) \nabla f(t,y)) \,d\mu_t(y) \,dt =
\iint \bigl((\partial_t f)(t,\varphi_t(x)) + a(t, \varphi_t(x)) (\nabla f)(t, \varphi_t(x))\bigr) \, d\mu(x) \,dt =
\iint \partial_t \bigl( f(t, \varphi_t(x)) \bigr) \,dt \,d\mu(x) = \int 0 \,d\mu = 0,
$$
where we have used the chain rule and the fact that $\partial_t\varphi_t(x) = a(t, \varphi_t(x))$ for a.e. $t$.

*Claim 1.* If $\varphi_t$ preserves the measure $\mu$ then $\operatorname{div} (a\mu) = 0$.

Indeed, if the flow of $\varphi_t$ preserves the measure $\mu$, i.e. $\mu_t = \mu$ for all $t$, then by (1)
$$
\operatorname{div} (a \mu) = 0. \tag{2}
$$
In particular, if $\mu$ is the Lebesgue measure then $\operatorname{div} a = 0$.

*Claim 2.* Suppose that $\mu_t$ is the *unique* solution of (2) with the initial condition $\mu_t|_{t=0} = \mu$ (e.g. in the class of non-negative measure-valued solutions, absolutely continuous with respect to Lebesgue measure). Then $\operatorname{div} (a\mu) = 0$ implies that the flow $\varphi_t$ preserves the measure $\mu$.

Indeed, uniqueness implies $\mu_t = \mu$ for a.e. $t$, that is $\mu$ is preserved by the flow of $a$.

(I learned this trick in some paper by E.O. Stepanov, but don't remember precisely which one.)