For the first question, consider $R = \mathbb{Z}[X]/(X^3 - 1)$, $P = (x+ 1)$ and $a = (x + 1)(1 + x + x^2)$ where $x$ denotes the image of $X$ in $R$.

In order to see that it provides us with a reduced one-dimensional Noetherian **counter-example**, note that $R/P \simeq \mathbb{Z}/2\mathbb{Z}$ and that $(x - 1)a = 0$. Hence $P = (2, x -1)$ is a maximal ideal of $R$ and $a$ is a zero-divisor. Any other prime ideal of $R$ containing $a$ has to contain $Q = (1 + x + x^2)$. As $x + 1$ doesn't divide $1 + x + x^2$ (since otherwise $1 + x + x^2$ would map to an even integer modulo $(x - 1)$), $Q$ is not a subset of $P$. Therefore $P$ is a minimal prime over $a$ which contains regular elements, namely $2$ and $x + 1$.

As the ideal $P$ is generated by the regular element $x + 1$, it is invertible, i.e., it satisfies $PP^{-1} = R \neq P$. It is easy to check that $P = ((a) :_R (1 + x + x^2))$, so that this counter-example satisfies also your extra requirements.

On the positive side, it is well-known that any minimal prime of a commutative ring with identity consists only of zero-divisors.
Thus the answer is yes if $a = 0$.

**Side note.**
In [1], D. Anderson and J. Pascual refers to the definition of *Property (A)*:
A commutative ring $R$ with identity enjoys property *(A)* if every faithful finitely generated ideal $I$ (i.e., $I$ is finitely generated and $(0:_R I) = 0$) is regular (i.e., $I$ contains at least one regular element). The authors remark that a ring in which $0$ has a primary decomposition satisfies Property $(A)$.
I am not sure if this can be helpful in your context, but if your ring $R$ has Property $(A)$, then $P$ consists solely of zero-divisors if and only if $P$ is not faithful, which may be easier to check.

[1] D. Anderson and J. Pascual, "Regular Ideals in Commutative Rings,
Sublattices of Regular Ideals, and Prüfer Rings", 1987.