Suppose $F$ is a formal group over $\mathbb{Z}_p$. There are few trivial condition that $f \in \mathbb{Z}_p[[t]]$, the power series representing the $p$-map should satisfy:
1)$f \equiv g(t^p) mod p$ for some $g \in \mathbb{Z}_p[[t]]$.
2)$f \equiv pt \mod t^2$.
Question: given an $f \in \mathbb{Z}_p[[t]]$ with the above conditions is there always a formal group $F$ over $\mathbb{Z}_p$ having $f$ as the $p$-map?
Even more trivial question: is $F$ determined by his $p$-map? If not is there a nice way to parametrize formal groups having the same p-map?
For your first question, if $p\ne2$, take $f(x)=px+x^{2p}$ and for $p=2$, take $f(x)=2x+x^6$. Neither can be the $[p]$-endomorphism of a formal group because, modulo $p$ they are $x^{2p}$ and $x^6$, respectively, while in characteristic $p$, the first nonzero term in a $[p]$-endom must have degree a power of $p$. So the requirement for $f$ to be a $[p]$-endomorphism is much stronger than just those two conditions you mention. It’s all in the encyclopedic, but I’m afraid difficult-to-read book on formal groups by Hazewinkel, a standard reference.
For your second question, yes, from any noninvertible endomorphism of a formal group $F$ you can recover the coefficients of $F$. This is one of the first results on formal groups in characteristic zero, and it’s true because you see from the equation $$ [p](F(x,y)=F\bigl([p](x),[p](y)\bigr) $$ that once you have the terms of $F$ of total degree less than $n$, the terms of degree $n$ are determined by the $x^n$-coefficient in the endomorphism. You should do some hand computations to see this. And the crucial part of the computation is not that the first-degree coefficient $c$ of $f$ is a nonunit, but that it is not a root of unity, because at just the right point of your computation, you need to divide by $c^n-c$.