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Here is an attempt at a Vandermonde-based proof.

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)}}{\sqrt{2!}},\ldots,\frac{\lambda^{\circ (k)}}{\sqrt{k!}},\ldots \end{pmatrix}\\\\ X &=& \begin{pmatrix} \mathbf{1}, x, \frac{x^{\circ (2)}}{\sqrt{2!}},\ldots,\frac{x^{\circ (k)}}{\sqrt{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.

Edit. Please look at the comments by Todd:

First note that,

1. We show that $L$ and $X$ are full-rank matrices
2. Each set of $n$ columns of $L$ (similarly $X$) is linearly independent
3. $M=LX^T$ is elementwise strictly positive.

However, as Todd has pointed out, the gap in the argument is that: the above points (1-2; I haven't thought about 3 yet) do not suffice to conclude that $LX^T$ has full rank. Since the conclusion is true (using Noam's or Darij's proofs), maybe there is a way to "rescue" the proof outline below---and if not, then I'll still let this "answer" hang in here to show an example of "what type of proof does not work for this problem!"

Here is an attempt at a Vandermonde-based proof.

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)}}{\sqrt{2!}},\ldots,\frac{\lambda^{\circ (k)}}{\sqrt{k!}},\ldots \end{pmatrix}\\\\ X &=& \begin{pmatrix} \mathbf{1}, x, \frac{x^{\circ (2)}}{\sqrt{2!}},\ldots,\frac{x^{\circ (k)}}{\sqrt{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.

Edit. Thanks to Todd for pointing out a possible mistake in my argument; until that is reconciled, I'll just leave the "answer" below more as an idea, than an answer.

Here is an attempt at a Vandermonde-based proof.

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)}}{\sqrt{2!}},\ldots,\frac{\lambda^{\circ (k)}}{\sqrt{k!}},\ldots \end{pmatrix}\\\\ X &=& \begin{pmatrix} \mathbf{1}, x, \frac{x^{\circ (2)}}{\sqrt{2!}},\ldots,\frac{x^{\circ (k)}}{\sqrt{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.

Here is an attempt at a Vandermonde-based proof.

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)}}{\sqrt{2!}},\ldots,\frac{\lambda^{\circ (k)}}{\sqrt{k!}},\ldots \end{pmatrix}\\\\ X &=& \begin{pmatrix} \mathbf{1}, x, \frac{x^{\circ (2)}}{\sqrt{2!}},\ldots,\frac{x^{\circ (k)}}{\sqrt{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.

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)}}{\sqrt{2!}},\ldots,\frac{\lambda^{\circ (k)}}{\sqrt{k!}},\ldots \end{pmatrix}\\\\ X &=& \begin{pmatrix} \mathbf{1}, x, \frac{x^{\circ (2)}}{\sqrt{2!}},\ldots,\frac{x^{\circ (k)}}{\sqrt{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.

Note: $LX^T$ is $n\times n$, and each entry of it is of the form: $1+\lambda_ix_j + \frac{1}{2!}{\lambda_i^2x_j^2} + \ldots = e^{\lambda_ix_j}$, so the product $LX^T$ is well-defined.

Edit. Thanks to Todd for pointing out a possible mistake in my argument; until that is reconciled, I'll just leave the "answer" below more as an idea, than an answer.

Here is an attempt at a Vandermonde-based proof.

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)}}{\sqrt{2!}},\ldots,\frac{\lambda^{\circ (k)}}{\sqrt{k!}},\ldots \end{pmatrix}\\\\ X &=& \begin{pmatrix} \mathbf{1}, x, \frac{x^{\circ (2)}}{\sqrt{2!}},\ldots,\frac{x^{\circ (k)}}{\sqrt{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.