Let $a > 0$. The equation $$(x-a) \, e^{x+a} = x $$ must be solved for $x > 0$. Since the solution does not have a closed form, I would like to obtain bounds for the solution.
Until now, I was able to show that $$ a < x < \sqrt{a}\sqrt{a + 1}. $$
The lower bound comes from $(x-a) e^{x+a} = x > 0$ and the upper bound comes from $e^{x+a} > 1+x+a$.
I'm interested in any better bounds than the ones I provided. However, to keep the question objective, I'm interested in the proof of the better upper bound $$x < \sqrt{a}\sqrt{a+e^{-a}},$$ which is very tight. I have tried to compare derivatives and use other inequalities but to no avail.
I also obtained the power series $$ x =\sqrt{a} - \frac{a}{4} + \frac{31 a^{3/2}}{96} +\frac{a^2}{16} - \frac{229 a^{5/2}}{92160} + \frac{a^3}{320} + \mathcal{O}(a^{7/2})$$ as $a \to 0^+$ for the solution; however, I'm after a simple bound.
Trying to solve for $x$ the equation $$x = (x-a)\, e^{(x+a)}$$ means that you are looking for the inverse of $$a=W\left(-x\,e^{-2 x}\right)+x \quad \text{for} \quad x>0\quad \text{and} \quad a>0 \tag 1$$
Consider the function $$f(x)=x -(x-a)\, e^{(x+a)}$$ Its first derivative cancels at $$x_*=W\left(e^{1-2 a}\right)+a-1$$ The second derivative test shows that this is a maximum. So, the solution is $x > x_*$.
Using Taylor series, we have $$f(x)=f(x_*)+\frac 12 f''(x_*)(x-x_*)^2+O\left((x-x_*)^3\right)$$ which gives a first estimate $$x_1=x_*+\sqrt{-2\frac{f(x_*)}{f''(x_*)}}$$
By Darboux theorem, we know that $x_1$ is an overestimate of the solution which will be reached without any overshoot using Newton method.
Similarly, $x=a$ is an underestimate of the solution.
Suppose that we make a series expansion around $x=a$
$$f(x)=a-\left(e^{2 a}-1\right)(x-a)-e^{2 a} \sum_{n=2}^\infty \frac {(x-a)^n}{(n-1)!}$$ Truncating to some low order, using power series reversion. gives $$x=a+\frac a {t-1}-\frac{a^2 t}{(t-1)^3}+\frac{a^3 t (3 t+1)}{2 (t-1)^5}+\cdots \tag 2$$$$ where $t=e^{2a}$.
This is again an overestimate of the solution.
Trying $(2)$ for a few values of $a$
$$\left( \begin{array}{ccc} a & \text{estimate} & \text{solution} \\ 1.00 & 1.136225395 & 1.134220030 \\ 1.25 & 1.350723894 & 1.350262505 \\ 1.50 & 1.572912849 & 1.572806865 \\ 1.75 & 1.801689832 & 1.801666058 \\ 2.00 & 2.035977718 & 2.035972557 \\ 2.25 & 2.274654840 & 2.274653759 \\ 2.50 & 2.516677007 & 2.516676788 \\ 2.75 & 2.761159048 & 2.761159005 \\ 3.00 & 3.007399649 & 3.007399641 \\ 3.25 & 3.254869728 & 3.254869728 \\ \end{array} \right)$$
We could do much better at the price of more terms in the initial series before truncation.
Edit
It seems that we can do better using the series expansion of $(1)$ around $x=a$ and making it the simplest $[1,1]$ padé approximant of it.
This would lead to a linear equation whose solution write $$x=a + 2a \frac{\alpha_0+\alpha_1 \,a}{\beta_0+\beta_1 \,a+\beta_2 \,a^2}$$ where $$\alpha_0=- w^2 (w+1)^2 \qquad \qquad \alpha_1=(w-1) w (w+1)^2$$
$$\beta_0=w^2\left(w^2+4 w+2\right)\qquad \beta_1=-4w\left( w^2-w-1\right)$$ $$\beta_2=2\left(w^3-w^2-w+1\right)$$ with $$w=W\left(-a e^{-2 a}\right)$$
Repeating the same calculations as above, there is a significant improvement for the small values of $a$.
$$\left( \begin{array}{ccc} a & \text{estimate} & \text{solution} \\ 1.00 & 1.134027194 & 1.134220027 \\ 1.25 & 1.350252492 & 1.350262505 \\ 1.50 & 1.572811420 & 1.572806865 \\ 1.75 & 1.801668349 & 1.801666058 \\ 2.00 & 2.035973280 & 2.035972557 \\ 2.25 & 2.274653949 & 2.274653759 \\ 2.50 & 2.516676833 & 2.516676788 \\ 2.75 & 2.761159015 & 2.761159005 \\ 3.00 & 3.007399643 & 3.007399643 \\ \end{array} \right)$$
With regard to the problem of $$x < \sqrt{a}\sqrt{a+e^{-a}}= b$$ we have $$f(b) <0 \qquad \text{and} \qquad f(b)\times f''(b) >0$$ So, by Darboux theorem, it is true.