Here is a Vandermonde-based proof (as an alternative to Noam's comment, which I did not immediately understand).
This proof below is an adaptation of (the proof of) Theorem 4.3.3 from this book by Bapat and Raghavan (their result is cast in terms of positive definite matrices).
Let $A=\lambda x^T$ (where $\lambda=(\lambda_1,\ldots,\lambda_n)$; likewise $x=(x_1,\ldots,x_n)$. and consider the Schur matrix $[e^{a_{ij}}]$. By direct expansion we have
\begin{equation*} [e^{a_{ij}}] = I + A + \frac{A^{(2)}}{2!} + \ldots + \frac{A^{(k)}}{k!} + \ldots \end{equation*} where $A^{(k)}$ is the Schur power of matrix $A$. We see that, \begin{equation*} A^{(2)} = (\lambda x^T) \circ (\lambda x^T) = (\lambda \circ \lambda)(x \circ x)^T. \end{equation*} Inductively, we obtain that $A^{(k)} = \lambda^{\circ (k)}x^{\circ (k)^T}$ (Schur powers), for $k=1,2,\ldots$.
Thus, it follows that \begin{equation*} [e^{a_{ij}}] = LX^T, \end{equation*} where $L$ and $X$ are infinite matrices with columns given by \begin{eqnarray*} L &=& \begin{pmatrix} \mathbf{1}, \lambda, \frac{\lambda^{\circ (2)}}{2!},\ldots,\frac{\lambda^{\circ (k)}}{k!},\ldots \end{pmatrix}\\\\ X &=& \begin{pmatrix} \mathbf{1}, x, \frac{x^{\circ (2)}}{2!},\ldots,\frac{x^{\circ (k)}}{k!},\ldots \end{pmatrix}, \end{eqnarray*} where $\mathbf{1}$ denotes the vector of all ones.
The desired invertibility of $[e^{a_{ij}}]$ will follow if we show that each of the matrices $L$ and $X$ has $n$ linearly independent columns. Since the $\lambda_i$ are distinct (given the ordering), as are the $x_i$, as per assumption, the Vandermonde matrix \begin{equation*} V = \begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1}\\\\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1}\\\\ \vdots & \vdots & \vdots & \vdots\\\\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \end{pmatrix} \end{equation*} is nonsingular, which shows already that the first $n$ columns of $X$ are linearly independent. A similar argument applies to $L$. Thus, their product $LX^T$ also has full rank, and its determinant is nonzero as desired.