Symplectic leaves of a Poisson-Lie group are solutions of the associated bivector field. 
Let $G$ be a Poisson-Lie group with the associated Poisson bivector field $\pi.$ Then it is well known that $\pi (e) = 0,$ where $e$ is the identity element in $G.$ Pavel Etingof mentioned in his lecture notes on compact quantum groups that $\{e\}$ is a symplectic leaf of $G.$

My question is $:$ How is symplectic leaf of a Poisson manifold defined?
Is it just the zeros of the bivector field associated with a Poisson manifold? In Chari and Pressley's book the definition of symplectic leaf is given in terms of an equivalence relation $\sim$ defined on the manifold such that $x \sim y$ if and only if there is a piecewise smooth path connecting $x$ and $y$ with each smooth piece of the path being a part of an integral curve of a Hamiltonian vector field defined on the underlying manifold. Could anyone please clarify it to me as to how are those two definitions equivalent? Any suggestion regarding this would be warmly appreciated.
Thanks for your time.
 A: Since the discussion in the comments got quite long, I summarize everything here.

Poisson bivector fields and the map $\pi^\sharp$
A Poisson bivector field $\pi\in\mathfrak{X}^2(M):=\Gamma(\wedge^2TM)$ on a manifold $M$ is a smooth map $\pi:M\to\wedge^2TM$ (and not $M\to T^*M$ that you said). There is an induced vector bundle morphism $\pi^\sharp:T^*M\to TM$ over the identity on $M$ which, on the level of sections, is defined via
$$ \langle \alpha, \pi^\sharp(\beta) \rangle = \pi(\alpha,\beta),\qquad \text{for all } \alpha,\beta\in \Omega^1(M).$$
In terms of exact $1$-forms $df$ with $f\in  C^\infty(M)$, this map is given by $\pi^\sharp(df)=\{f,\cdot\}\in\mathfrak X(M)$; the Hamiltonian vector field of the function $f$ is defined to be that vector field $X_f:= \pi^\sharp(df)$.
Hamiltonian distribution and symplectic leaves
The map $\pi^\sharp$ constructed above gives the (not necessarily regular) smooth distribution of the manifold $M$ spanned by the Hamiltonian vector fields. This distribution is closed under the Lie bracket of vector fields on $M$ due to one of the most fundamental identities of Poisson manifolds: $[X_f,X_g]=X_{\{f,g\}}$ for all smooth functions $f,g$ on $M$; that is, the distribution spanned by the Hamiltonian vector fields on $M$ is integrable, and hence can be integrated to a foliation on $M$. The leaves of this distribution are called the symplectic leaves of the Poisson manifold $M$. Note that, by definition, for any leaf $S$ passing through the point $x\in M$ we have $T_x S= \pi^\sharp_x(T_x^*M)\subset T_xM$.
In terms of the local Weinstein splitting, given the Darboux-Weinstein coordinates $(U,q^i,p^i,y^k)$ around a point $x$, where
$$\pi|_U=\sum_i\frac{\partial}{\partial q^i}\wedge \frac{\partial}{\partial p^i} + \sum_{k,\ell} \pi’_{kl}(y) \frac{\partial}{\partial y^k} \wedge \frac{\partial}{\partial y^l}$$
with $\pi’(x)=0$, the symplectic leaf is described by the equations $y^k=0$ for all $k$.
Another more differential geometric way to view these leaves are, as you said, all the points that can be connected by pieces of flow curves of Hamiltonian vector fields, since they are precisely the submanifolds obtained by the span of all the Hamiltonian vector fields and, as a foliation, they are always connected.
More precisely, in the Darboux-Weinstein coordinates around $x\in M$ described above we can see that $X_{q^i}=\frac{\partial}{\partial p^i}$ and $X_{p^i}=-\frac{\partial}{\partial q^i}$, which means that the lines parallel to the axes are Hamiltonian flows. So we can connect any point in $S$ with the origin which corresponds to $x$ by projecting it onto the $q$-axis and then moving along the axis until we reach the origin. In the general case where two points are not in the same Darboux-Weinstein chart, we can connect them by finitely many charts of this form and proceed inductively.
Note that this idea has nothing to do with the Poisson structure, you can find more about it by looking up foliations/distributions on manifolds and the Frobenius theorem for regular distributions. For singular distributions one needs the Stefan-Sussmann theorem; you can find a few things here, but it's still not a complete theory. I'm afraid I don't know of any elementary text for this. The only things I could suggest are either the original papers by Stefan and Sussmann (given in the Wikipedia article for singular distributions that I linked), or this paper (Corollary 3.14 and the comments right after that) from which I also learned about it when I was a student and it is Poisson-geometry-oriented.
Zeros of the Poisson structure
Suppose that $x\in M$ is a zero of the Poisson bivector field, i.e. $\pi_x=0$. Then the tangent space of the leaf $S$ passing through $x$ is given by $T_x S=\pi^\sharp_x(T_x^*M)=\{0\}$. This implies that the dimension of the manifold $S$ is $0$, i.e. $S$ is a point. Since $x\in S$ we get that $S=\{x\}$.
In your case of the Poisson-Lie group $G$, this is applied to the identity element $e$ which is always a zero of its Poisson bivector field. And so $\{e\}$ is a symplectic leaf of $G$.
