The trick is that you can look at just the highest-degree terms.
Note that $$e_1^{k_1}\dots e_n^{k_n}\cdot e_1^{j_1}\dots e_n^{j_n}=e^{k_1+j_1}\dots e_n^{k_n+j_n}+\text{lower degree terms}$$ since all the other terms come from replacing two of the factors by their bracket and thus have lower degree. So the highest degree part of $x\cdot y$ is just obtained by multiplying the highest degree parts of $x$ and $y$ as if the $e_i$ all commuted (i.e., treating them as variables in an ordinary polynomial ring). Since a polynomial ring in $n$ variables has no zero divisors, the highest degree part of $x\cdot y$ is nonzero (assuming $x$ and $y$ are nonzero) and thus $x\cdot y$ is nonzero.
Or, without picking a basis, if $x$ has degree $d$ and $y$ has degree $e$, we look at the image $x'$ of $x$ in $A_d$ and the image $y'$ of $y$ in $A_e$ where $A$ is the associated graded algebra of $U(\mathfrak{g})$. By definition of the ring structure on $A$, $x'\cdot y'$ is the image of $xy$ in $A_{d+e}$. If $A$ has no zero divisors and $x$ and $y$ are nonzero, then $x'$ and $y'$ are nonzero and thus $x'y'$ is nonzero, so $xy$ must be nonzero. (This argument shows more generally that if the associated graded ring of a filtered ring has no zero divisors, then neither does the original ring.)
$\DeclareMathOperator{\tr}{tr}\DeclareMathOperator{\ad}{ad}$I will expand my comment here. I am assuming (though I am not sure if this is the case) that the OP is interested in real Lie algebras. If $\mathfrak{g}$ is the Lie algebra of a compact semisimple Lie group, then with respect to any orthonormal basis of $\mathfrak{g}$ with respect to (minus) the Killing form, the matrix representing $\operatorname{ad}_x \in \mathfrak{gl}(\mathfrak{g})$ is skew-symmetric, for any $x \in \mathfrak{g}$.
Define the Killing form $B(-,-): \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ by:
$$B(x,y) = \tr(\ad_x \circ \ad_y).$$
I claim that for any $z \in \mathfrak{g}$, $B([z,x],y) + B(x,[z,y]) = 0$. For short, it is common to say that the Killing form is $\ad$-invariant.
$$ \begin{align*} & B([z,x],y) + B(x,[z,y]) \\
= & \,\tr(\ad_{[z,x]} \circ \ad_y)+
\tr(\ad_x,\ad_{[z,y]}) \\
= & \,\tr([\ad_z,\ad_x] \circ \ad_y) + \tr(\ad_x \circ \, [\ad_z, \ad_y]) \\
= & \,0 \end{align*}$$
by the cyclic property of the trace ($\tr(BCA) = \tr(ABC)$). I have also used the fact that $\ad_{[x,y]} = [\ad_x,\ad_y]$.
There is another fact from Lie theory that we need, namely that if $\mathfrak{g}$ is the Lie algebra of a compact semisimple Lie group, then its Killing form is negative definite.
Let us now choose an orthonormal basis of $\mathfrak{g}$, say $x_1,\ldots,x_n$, with respect to minus the Killing form. Note that, for any $x \in \mathfrak{g}$, we have
$$ -B(\ad(x)(x_i),x_j) = -B([x,x_i],x_j) = B(x_i, [x,x_j]) = B(x_i, \ad_x(x_j)).$$
Hence the matrix representing $\ad_x$ with respect to the orthonormal basis $x_1,\ldots,x_n$ is skew-symmetric.
Best Answer
The Lie algebra $\mathfrak{gl}_2=\langle e_{11},e_{12},e_{21},e_{22}\rangle$ is a counterexample. If $e_{12}<e_{21}$, then by multiplication, $e_{12}<e_{21}<e_{11},e_{22}$, but then $[e_{12},e_{22}]=e_{22}$ does not satisfy (1). If $e_{21}<e_{12}$, then by multiplication, $e_{21}<e_{12}<e_{11},e_{22}$, but then $[e_{21},e_{11}]=e_{21}$ does not satisfy (1).
Using @darijgrinberg's comments, I've found on Wiki, that $\frak{g}$ is solvable iff there is a finite sequence of subalgebras $\mathfrak{a}_{i}$ of $\mathfrak{g}$, such that: $$\mathfrak{g}=\mathfrak{a}_0\supset \mathfrak{a}_1\supset ...\supset \mathfrak{a}_m=0,\quad \dim \mathfrak{a}_i/\mathfrak{a}_{i+1}=1,\;\; \mathfrak{a}_{i+1}\text{ is an ideal in }\mathfrak {a}_{i},\;\;\forall i.$$ This corresponds precisely to my condition (1): $e_i$ is the basis element of $\mathfrak{g}_i/\mathfrak{g}_{i+1}$.