Relationships between the roots of an entire function and the roots of its derivative

I don't know if there are any general results about these, but when $f$ is a polynomial, these must be in essence results about symmetric functions. If $f(z)=z^n+a_{n-1} z^{n-1}+\cdots+a_1z+a_0$ then $Z(1)=-a_1/a_0$, $Z(2)=Z(1)^2-2a_2/a_0$ etc. In this case your results surely specialize to polynomial identities in the $a_j$.

Let $f(x) = (1 - r_1 x)...(1 - r_n x)$ be a polynomial. Then $f(x) = 1 - e_1 x^1 + e_2 x^2 \mp ... $ where the $e_i$ are the elementary symmetric functions in the $r_i$. We define also $p_k = \sum_i r_i^k$, the power symmetric functions in the $r_i$. Then Newton's identities state that

$$ke_k = \sum_{i=1}^{k} (-1)^{i-1} e_{k-i} p_i.$$

This identity is equivalent to the generating function identity

$$\frac{f'(x)}{f(x)} = \sum_{i=1}^{n} \frac{r_i}{1 - r_i x} = \sum_{k \ge 0} p_{k+1} x^k$$

which follows from taking logarithmic derivatives on both sides. Now, you want to relate the functions $p_k$ to the functions $\tilde{p}_k$, the power symmetric functions of the reciprocals of the roots of the derivative $f'(x)$. Applying Newton's identities to $f'(x) = -e_1 + 2e_2 x - 3e_3 x^2 \pm ...$ gives

$$k(k+1) \frac{e_{k+1}}{e_1} = \sum_{i=1}^k (-1)^{i-1} \frac{(k+1-i) e_{k+1-i}}{e_1} \tilde{p}_i.$$

This pair of identities should get you your results (at least formally, letting $n \to \infty$). You may or may not find it useful to write Newton's identities in the equivalent form

$$e_n = \frac{1}{n!} \sum_{\sigma \in S_n} \text{sgn}(\sigma) p_{\sigma}$$

where $p_{\sigma} = p_{\lambda_1} ... p_{\lambda_i}$, where $\sigma$ has cycle type $(\lambda_1, ... \lambda_i)$. This gives

$$\frac{(n+1) e_{n+1}}{e_1} = \frac{1}{n!} \sum_{\sigma \in S_n} \text{sgn}(\sigma) \tilde{p}_{\sigma}.$$