We can use a procedure known as "bootstrapping" to determine an approximation for the Lambert $W$ function. Let's go back to its definition.
For $x > 0$ the equation
$$
we^w = x
$$
has exactly one positive solution $w = W(x)$ which increases with $x$. Note that $(w,x) = (1,e)$ is one such solution, so if $x > e$ then $w > 1$. By taking logarithms of both sides of the equation we get
$$
\log w + w = \log x
$$
or
$$
w = \log x - \log w. \tag{1}
$$
When $x > e$ we therefore have
$$
w = \log x - \log w < \log x.
$$
In other words, our first approximation is that
$$
1 < w < \log x \tag{2}
$$
when $x > e$. We then have
$$
0 < \log w < \log\log x,
$$
and plugging this into $(1)$ yields
$$
\log x - \log \log x < w < \log x, \tag{3}
$$
where the left side is positive for $x > 1$. Taking logarithms as before yields
$$
\log\log x + \log\left(1 - \frac{\log\log x}{\log x}\right) < \log w < \log\log x,
$$
and upon substituting this back into $(1)$ we get
$$
\log x - \log\log x < w < \log x - \log\log x - \log\left(1 - \frac{\log\log x}{\log x}\right).
$$
Since $w = W(x)$ we have shown that
$$
\log x - \log\log x < W(x) < \log x - \log\log x - \log\left(1 - \frac{\log\log x}{\log x}\right) \tag{4}
$$
for $x > e$.
In your particular case we're interested in $W(e^{x+a})$, for which we have
$$
x+a - \log(x+a) < W(e^{x+a}) < x+a - \log(x+a) - \log\left(1 - \frac{\log(x+a)}{x+a}\right)
$$
for $x+a > 1$. In this sense we have
$$
W(e^{x+a}) \approx x+a - \log(x+a) = x\left(1 - \frac{\log(x+a) - a}{x}\right) \tag{5}
$$
when $x+a$ is large. Now by applying Taylor series a couple times we see that, for $x$ large and $a \ll x$,
$$
\begin{align}
\frac{\log x - a}{x+1} &= \frac{\log x - a}{x} \cdot \frac{1}{1+\frac{1}{x}} \\
&\approx \frac{\log x - a}{x} \left(1-\frac{1}{x}\right) \\
&= \frac{\log x - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a-a) - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a) + \log\left(1-\frac{a}{x+a}\right) - a}{x} - \frac{\log x - a}{x^2} \\
&= \frac{\log(x+a) - a}{x} + \frac{\log\left(1-\frac{a}{x+a}\right)}{x} - \frac{\log x - a}{x^2} \\
&\approx \frac{\log(x+a) - a}{x} - \frac{a}{x(x+a)} - \frac{\log x - a}{x^2} \\
&\approx \frac{\log(x+a) - a}{x}.
\end{align}
$$
We may then conclude from $(5)$ that
$$
W(e^{x+a}) \approx x \left(1 - \frac{\log x - a}{x+1}\right)
$$
for $x$ large and $a \ll x$.
Best Answer
There already were methods mentioned to find $x=e$, but for the other root, apply Lagrange reversion: $$\frac {x\ln x}{\ln x+1}=\frac e2\mathop\iff^{x=\sqrt[w]e} e(w+1)=2\sqrt[w]e\implies x=\frac1e-\sum_{n=1}^\infty\frac{\left(\frac 2e\right)^n}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}\frac{e^\frac{n+1}w}{w^2}\right|_{-1}$$
Using a Maclaurin series and factorial power $n^{(m)}$ for the derivatives:
$$\left.\frac{d^{n-1}}{dw^{n-1}}\frac{e^\frac{n+1}w}{w^2}\right|_{-1}=\sum_{m=0}^\infty\frac{(-n-1)^m(-m-2)^{(n-1)}}{m!}$$
This uses a confluent hypergeometric function. We rewrite it as a Laguerre polynomial sum and apply a Laguerre $\operatorname L_n^1(x)$ integral representation,
$$\bbox[4px,border: 2.7px double #5ADEFF]{\frac {x\ln x}{\ln x+1}=\frac e2\implies x=e,x=1-\sum_{n=1}^\infty\left(\frac2e\right)^{n-1}\,_1\text F_1(n;2;-n)=\frac1e+\frac1{2\pi}\int_0^{2\pi}e^{it-e^{it}-1}\ln(1-2(e^{it}+1)e^{-e^{it}-it-2})dt}$$
$x=0.28285\dots$ is shown here