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][1] (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.

  [1]: http://books.google.com/books?id=J-u2t23tkG4C&source=gbs_navlinks_s