1. **State the problem:** Solve for $x$ in terms of $a$ and $b$ from the equation $$\frac{a^{2}b}{x^{2}} + \left(1 + \frac{b}{x}\right) a = 2b + \frac{a^{2}}{x}.$$\n\n2. **Rewrite the equation:** Distribute $a$ in the second term on the left side:\n$$\frac{a^{2}b}{x^{2}} + a + \frac{ab}{x} = 2b + \frac{a^{2}}{x}.$$\n\n3. **Bring all terms to one side:**\n$$\frac{a^{2}b}{x^{2}} + a + \frac{ab}{x} - 2b - \frac{a^{2}}{x} = 0.$$\n\n4. **Group terms by powers of $\frac{1}{x}$:**\n$$\frac{a^{2}b}{x^{2}} + \left(\frac{ab}{x} - \frac{a^{2}}{x}\right) + (a - 2b) = 0.$$\n\n5. **Simplify the terms inside the parentheses:**\n$$\frac{a^{2}b}{x^{2}} + \frac{ab - a^{2}}{x} + (a - 2b) = 0.$$\n\n6. **Let $y = \frac{1}{x}$ to simplify:**\n$$a^{2}b y^{2} + (ab - a^{2}) y + (a - 2b) = 0.$$\n\n7. **This is a quadratic equation in $y$:**\n$$A = a^{2}b, \quad B = ab - a^{2}, \quad C = a - 2b.$$\n\n8. **Use the quadratic formula:**\n$$y = \frac{-B \pm \sqrt{B^{2} - 4AC}}{2A} = \frac{-(ab - a^{2}) \pm \sqrt{(ab - a^{2})^{2} - 4 a^{2}b (a - 2b)}}{2 a^{2} b}.$$\n\n9. **Simplify numerator:**\n$$y = \frac{a^{2} - ab \pm \sqrt{(ab - a^{2})^{2} - 4 a^{2} b (a - 2b)}}{2 a^{2} b}.$$\n\n10. **Recall $y = \frac{1}{x}$, so:**\n$$x = \frac{1}{y} = \frac{2 a^{2} b}{a^{2} - ab \pm \sqrt{(ab - a^{2})^{2} - 4 a^{2} b (a - 2b)}}.$$\n\n**Final answer:**\n$$\boxed{x = \frac{2 a^{2} b}{a^{2} - ab \pm \sqrt{(ab - a^{2})^{2} - 4 a^{2} b (a - 2b)}}}.$$