What you are trying to do is done in complete detail in Leclerc's paper "Dual canonical bases, quantum shuffles and q-characters" based on the paper "Standard Lyndon bases of Lie algebras and enveloping algebras" by Lalonde and Ram.

The root spaces are one dimensional, so you have to put conditions on $\alpha$ and $\beta$ to get a well defined $E_{\alpha+\beta}$. Roughly speaking, you start by observing that convex orderings on positive roots are equivalent to total orderings on simple roots. So fix one such ordering.

In the (quantized) enveloping algebra you can start to think about words in the simple roots associated to (ordered) products of the corresponding $E_i$. For reasons explained in these papers, the monomial $E_{i_1}\cdots E_{i_n}$ corresponds to the linear combination of words obtained by (quantum) shuffling the symbols $i_1,i_2,\ldots,i_n$ together in all possible ways and summing the outcomes (Note that in the quantum context, this is a noncommutative product!). Write this shuffle product as $$i_1*\cdots*i_n=\sum_{w\in S_n}q^{(\star)}\; i_{w(1)}\cdots i_{w(n)},$$ where $(\star)$ can be given explicitly. Let $\max(i_1*\cdots*i_n)$ be the largest monomial (in the lexicographic order) in the sum $i_1*\cdots*i_n$ (such a monomial is called \emph{good} or \emph{standard}). The set of all possible $\max(i_1*\cdots*i_n)$ for $n\geq 1$ can be identified with the crystal $B(\infty)$. It is also worth pointing out that there is a nice triangularity property: $E_{i_1}E_{i_2}\cdots E_{i_n}$ corresponds to $$i_1*i_2*\cdots*i_n=i_1i_2\cdots i_n+\sum(words < i_1i_2\cdots i_n).$$

Anyway, given a word $w\in \mathcal{B}(\infty)$, call it "standard Lyndon" if it is larger than any subword ($w=w_1 w_2$ implies $w_1 < w$). The key fact is that the set of standard Lyndon words are in bijection with the set of positive roots. Leclerc explains how to calculate the set standard Lyndon words in Proposition 25 of his paper.

For a positive root $\alpha$, let $w(\alpha)$ be the corresponding word. You construct the root vectors inductively by $E_\alpha=[E_{\alpha_1},E_{\alpha_2}]$, where

$\bullet$ $w(\alpha)=w(\alpha_1)w(\alpha_2)$, $\alpha=\alpha_1+\alpha_2$, and $\alpha_1\neq 0$, $\alpha_2\neq0$ are positive roots,

$\bullet$ $w(\alpha_1)$ is standard Lyndon, and

$\bullet$ if $w(\alpha)=w_1w_2$, with $w_1$ standard Lyndon, then $w_1$ is a shorter word than $w(\alpha_1)$.

This construction has also been carried out in the affine case by Beck-Chari-Pressley and Beck-Nakajima. In my paper with Melvin and Mondragon, we make these calculations explicitly in all types (well, type $E_8$ isn't exactly explicit, but almost). Note, however, we use the middle east reading convention when defining the lex. ordering.

What you are trying to do is done in complete detail in Leclerc's paper "Dual canonical bases, quantum shuffles and q-characters" (MR) based on the paper "Standard Lyndon bases of Lie algebras and enveloping algebras" (MR) by Lalonde and Ram.

The root spaces are one dimensional, so you have to put conditions on $\alpha$ and $\beta$ to get a well defined $E_{\alpha+\beta}$. Roughly speaking, you start by observing that convex orderings on positive roots are equivalent to total orderings on simple roots. So fix one such ordering.

In the (quantized) enveloping algebra you can start to think about words in the simple roots associated to (ordered) products of the corresponding $E_i$. For reasons explained in these papers, the monomial $E_{i_1}\cdots E_{i_n}$ corresponds to the linear combination of words obtained by (quantum) shuffling the symbols $i_1,i_2,\ldots,i_n$ together in all possible ways and summing the outcomes (Note that in the quantum context, this is a noncommutative product!). Write this shuffle product as $$i_1*\cdots*i_n=\sum_{w\in S_n}q^{(\star)}\; i_{w(1)}\cdots i_{w(n)},$$ where $(\star)$ can be given explicitly. Let $\max(i_1*\cdots*i_n)$ be the largest monomial (in the lexicographic order) in the sum $i_1*\cdots*i_n$ (such a monomial is called \emph{good} or \emph{standard}). The set of all possible $\max(i_1*\cdots*i_n)$ for $n\geq 1$ can be identified with the crystal $B(\infty)$. It is also worth pointing out that there is a nice triangularity property: $E_{i_1}E_{i_2}\cdots E_{i_n}$ corresponds to $$i_1*i_2*\cdots*i_n=i_1i_2\cdots i_n+\sum(words < i_1i_2\cdots i_n).$$

Anyway, given a word $w\in \mathcal{B}(\infty)$, call it "standard Lyndon" if it is larger than any subword ($w=w_1 w_2$ implies $w_1 < w$). The key fact is that the set of standard Lyndon words are in bijection with the set of positive roots. Leclerc explains how to calculate the set standard Lyndon words in Proposition 25 of his paper.

For a positive root $\alpha$, let $w(\alpha)$ be the corresponding word. You construct the root vectors inductively by $E_\alpha=[E_{\alpha_1},E_{\alpha_2}]$, where

$\bullet$ $w(\alpha)=w(\alpha_1)w(\alpha_2)$, $\alpha=\alpha_1+\alpha_2$, and $\alpha_1\neq 0$, $\alpha_2\neq0$ are positive roots,

$\bullet$ $w(\alpha_1)$ is standard Lyndon, and

$\bullet$ if $w(\alpha)=w_1w_2$, with $w_1$ standard Lyndon, then $w_1$ is a shorter word than $w(\alpha_1)$.

This construction has also been carried out in the affine case by Beck–Chari–Pressley (MR) and Beck–Nakajima (MR). In my paper with Melvin and Mondragon (MR), we make these calculations explicitly in all types (well, type $E_8$ isn't exactly explicit, but almost). Note, however, we use the middle east reading convention when defining the lex. ordering.