Let me elaborate on reuns' idea and show how one can find coefficients $a_i$ by solving a certain ODE.

Let's define: $$f(z) := \sum_{i\geq 3} a_i z^{i-3}$$ so that we get a functional equation: $$\mathrm{li}\big( -x(\frac{1}{t}+tf(t)) \big) = x,$$ where $t=t(x):=-\frac{1}{W_1(-e/x)}$. Differentiating this equation with respect to $x$, and then substituting $x=\frac{et}{\exp(-1/t)}$, we obtain a differential equation: $$(\star)\qquad t^3f'(t) - t(1-2t)f(t) - (1-t)\log(1-t^2f(t)) + t = 0$$ with the initial condition $f(0)=1$.

I've played with ODE $(\star)$ in Maple. In principle, Maple can solve it in the following form: $$f(t) = \frac{1-\exp\big(r-1/t\big)}{t^2},$$ where $r$ is the root of the equation: $\mathrm{Ei}(1,-1-r)=-\frac{et}{\exp(-1/t)}$. However, I'm not sure if this implicit form is that useful.

On the other hand, Maple can solve $(\star)$ in power series of given order, thus computing the coefficients $a_i$. For example,

Order:=15: dsolve( { t^3*diff(f(t),t) - t*(1-2*t)*f(t) - (1-t)*log(1-t^2*f(t)) + t = 0, f(0)=1 }, f(t), series);

gives

$$f \left( t \right) =1+3\,t+11\,{t}^{2}+{\frac{105}{2}}{t}^{3}+{\frac{613}{2}}{t}^{4}+{\frac{12635}{6}}{t}^{5}+{\frac{99677}{6}}{t}^{6}+{\frac{1774391}{12}}{t}^{7}+{\frac{17582819}{12}}{t}^{8}+{\frac{1919343719}{120}}{t}^{9}+{\frac{22882040099}{120}}{t}^{10}+{\frac {295793507053}{120}}{t}^{11}+{\frac{1373607474819}{40}}{t}^{12}+{\frac{323119030735871}{630}}{t}^{13}+{\frac{20600974525589671}{2520}}{t}^{14}+O \left( {t}^{15} \right).$$

Let me elaborate on reuns' idea and show how one can find coefficients $a_i$ by solving a certain ODE.

Let's define: $$f(z) := \sum_{i\geq 3} a_i z^{i-3}$$ so that we get a functional equation: $$\mathrm{li}\big( -x(\frac{1}{t}+tf(t)) \big) = x,$$ where $t=t(x):=-\frac{1}{W_1(-e/x)}$. Differentiating this equation with respect to $x$, and then substituting $x=\frac{et}{\exp(-1/t)}$, we obtain a differential equation: $$(\star)\qquad t^3f'(t) - t(1-2t)f(t) - (1-t)\log(1-t^2f(t)) + t = 0$$ with the initial condition $f(0)=1$.

I've played with ODE $(\star)$ in Maple. In principle, Maple can solve it in the following form: $$f(t) = \frac{1-\exp\big(r-1/t\big)}{t^2},$$ where $r$ is the root of the equation: $\mathrm{Ei}(1,-1-r)=-\frac{et}{\exp(-1/t)}$. However, I'm not sure if this implicit form is that useful.

On the other hand, Maple can solve $(\star)$ in power series of given order, thus computing the coefficients $a_i$. For example,

Order:=15: dsolve( { t^3*diff(f(t),t) - t*(1-2*t)*f(t) - (1-t)*log(1-t^2*f(t)) + t = 0, f(0)=1 }, f(t), series);

gives

$$f \left( t \right) =1+3\,t+11\,{t}^{2}+{\frac{105}{2}}{t}^{3}+{\frac{613}{2}}{t}^{4}+{\frac{12635}{6}}{t}^{5}+{\frac{99677}{6}}{t}^{6}+{\frac{1774391}{12}}{t}^{7}+{\frac{17582819}{12}}{t}^{8}+{\frac{1919343719}{120}}{t}^{9}+{\frac{22882040099}{120}}{t}^{10}+{\frac {295793507053}{120}}{t}^{11}+{\frac{1373607474819}{40}}{t}^{12}+{\frac{323119030735871}{630}}{t}^{13}+{\frac{20600974525589671}{2520}}{t}^{14}+O \left( {t}^{15} \right).$$

I've added the sequences of coefficients numerators / denominators to the OEIS as A337734 and A337735, respectively.