One answer is "no." Another is, "of course."
No
To simplify notation, let $\lambda(x) = 1/(1 + \exp(-x))$, the inverse logit. Because $\lambda(x) = 1 - \lambda(-x)$,
$$\beta_0 + \beta_1 \lambda(x) = (\beta_0 + \beta_1) - \beta_1 \lambda(-x)).$$
Therefore it is impossible to distinguish the parameters $(\beta_0, \beta_1, \beta_2, \beta_3)$ from $(\beta_0+\beta_1, -\beta_1, -\beta_2, -\beta_3)$.
Of course
Let us stipulate that the first nonzero element of $(\beta_1, \beta_2, \beta_3)$ must be positive. That resolves the indeterminacy. We still need a model for the errors. If we suppose, for instance, that $Y - \left(\beta_0 + \beta_1 \lambda(\beta_2 X_2 + \beta_3 X_3)\right)$ has a Normal distribution and the various $Y$'s are independent, then we can use least squares to estimate the parameters. There is no exact solution to this nonlinear optimization problem, but it is straightforward to do numerically.
This graphic shows 50 points generated with standard Normal values for $X_1$ and $X_2$, parameter $\beta = (1,2,1/2,-1)$, with iid Normal errors of standard deviation 1/2. The surface shows the fit, $\hat{\beta} = (2.68, -1.23, -0.89, 1.75) \sim (1.45, 1.23, 0.89, -1.75)$.
Least squares is the maximum likelihood with iid Normal errors. With another error distribution, use MLE directly. You can obtain asymptotic confidence intervals for the parameters in the standard ways.
Expit
="Inverse logit function" from the msm package? $\endgroup$