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 $a \ll x$.