Is it possible to find a function $x:[0,T]\to [0,x_0]$ such that, for a fixed $0<\lambda<1$ we have:
$$\dfrac{1}{1+\lambda}\left (1-\dfrac{x(t)}{x_0}\right )^{1+\lambda} +\dfrac{1}{1-\lambda} \left (1-\dfrac{x(t)}{x_0}\right )^{1-\lambda}=\dfrac{2}{1-\lambda^2}-\dfrac{2v_P t}{x_0},\ \forall t\in I$$
At least, there is such a function? Is it unique?
This equation gives a pursuit curve. See Bouguer Problem, on Chases and Escapes: The Mathematics of Pursuit and Evasion by Paul Nahin, page 7.
Note that for $t=T$ your equation gives $\frac 1 {1-\lambda^2} = \frac {v_P T} {x_0}$. If this is not satisfied, there can be no solution as you wish. Let us call this condition "the existence condition".
Consider the function $f:[0,1] \to \Bbb R$, $f(y) = \frac 1 {1-\lambda} y^{1-\lambda} + \frac 1 {1+\lambda} y^{1+\lambda}$. Note that $f$ is derivable and $f'(y) = y^{-\lambda} + y^{\lambda} >0$, so $f$ is strictly increasing, therefore its range will be $[0, \frac 2 {1-\lambda ^2}]$, and $f$ will admit an inverse $g:[0, \frac 2 {1-\lambda ^2}] \to [0,1]$. Since $f'$ is continuously derivable and $f'(y) \ne 0 \space \forall y \in [0,1]$, by the inverse function theorem, $g$ will also be continuously derivable.
Your equation can be rewritten as $f(1 - \frac {x(t)} {x_0}) = \frac 2 {1-\lambda ^2} - \frac {2 v_P t} {x_0}$. Since the existence condition is satisfied, the left-hand side is in $[0, \frac 2 {1-\lambda^2}]$ and we can apply $g$, so $1 - \frac {x(t)} {x_0} = g(\frac {2 v_P t} {x_0})$, which means that $x(t) = x_0 - x_0 g(\frac 2 {1-\lambda ^2} - \frac {2 v_P t} {x_0})$. Note that $x(0) = 0$ and, thanks to the existence condition, $x(T) = x_0$. Furthermore, since $g$ is continuously derivable, so will be $x$.
We can conclude that the existence condition is necessary and sufficient for your question to be answered in the affirmative.