It seems, experimentally, that the set of nyldon words is a Hall set, with the same construction of the basis as for Lyndon words, that is if $w=uv$ where $w,u,v$ are nyldon words and $|v|$ is maximal, then $\pi(w)= \pi(u) \pi(v)$ (note that, by definition, we have $u>_{lex} v$). To show that this mapping gives a Hall set, it suffice to show (I think, I am not an expert on Hall set...) that (1) $w>_{lex}v$, (2) $u>_{lex}v$ and (3) if $u$ is not a single letter then $y\le_{lex} v$, where $u=xy$, $x,y$ are nyldon words and $|y|$ maximal. (2) is true by definition, so (1) is true. Maybe we can show (3) using the maximality of $|v|$...
Moreover, it seems that we have the same properties for other total orderings on words: "shortlex" (i.e. $u<v$ if $|u|<|v|$, or $|u|=|v|$ and $u<_{lex} v$) and (say) "shortrevlex" (i.e. $u<v$ if $|u|<|v|$, or $|u|=|v|$ and $u>_{lex} v$).
More strangely, Conjecture 1 and 2 seem true also for "longlex" ($u<v$ if $|u|>|v|$, or $|u|=|v|$ and $u<_{lex} v$) and "longrevlex" ($u<v$ if $|u|>|v|$, or $|u|=|v|$ and $u>_{lex} v$), but the mapping $\pi(w)= \pi(u) \pi(v)$ with $w=uv$ and $|v|$ maximal do not gives a Hall set... (And neiter if $|v|$ is minimal instead of maximal.)
edit:
I quickly look in "Algebres de Lie Libres et Monoides Libres" from Viennot, and some things are more clear now.
Nyldon words seem to be a Lazard right factorization. I.e. the following procedure seems to give all Nyldon words of size at most n:
$Y_0 := \{0,1\}$ and $i:=0$
While $Y_i$ has words of size at most n :
$u_{i+1}$ = $\min_{lex} \{ y : y \in Y_i \text{ s.t } | y | <= n\}$
$Y_{i+1}$ = $( Y_i \setminus \{ u_{i+1}\} ) u_i^*$
$i:=i+1$
Eg for $n=4$ $Y_0 : \{0, 1\}$, $u_1 : 0 $, $Y_1 : \{1, 10, 100, 1000, ...\}$, $u_2 : 1 $, $Y_2 : \{10, 100, 1000, 1001, 101, 1011, ...\}$, $u_3 : 10 $, $Y_3 : \{100, 1000, 1001, 101, 1011, ...\}$, $u_4 : 100 $, $Y_4 : \{1000, 1001, 101, 1011, ...\}$, $u_5 : 1000 $, $Y_5 : \{1001, 101, 1011, ...\}$, $u_6 : 1001 $, $Y_6 : \{101, 1011, ...\}$, $u_7 : 101 $, $Y_7 : \{1011, ...\}$, $u_8 : 1011$
(where ... represents words of size >n)
This is not the usual Lazard (left) factorization definition, but it is if we take the mirror.
(Note: Lyndon words are left and right Lazard factorizations, thus are called "regular factorization" in [Viennot]. Nyldon words are not a left Lazard factorization, so I think it is difficult to find direct correspondence between Lyndon and Nyldon words...)
Let $U_n$ be the output (the set of all $u_i$) of the procedure with a fixed n, and $U=\cup_{n\in N} U_n$. It is known that $U$ is a factorization, since it is a Lazard elimination procedure, thus Conjectures 1 and 2 are true for the set $U$.
Now it suffice to show that $U$ are exactly Nyldon words, thus U_n are Nyldon words of size <=n. Maybe we can follow the following way. (Note that the procedure gives the set U_n in the lexicographic order.)
For every proper suffix s of $u_i$, $s \cap Y_i = \emptyset$
For every proper suffix s of $u_i$, $s \cap Y_i^* = \emptyset$
For every $w\in U$ and s suffix of w, if $s\in U$ then $s<_{lex} w$
By induction, $U_n$ is a set of Nyldon words.
