= Solution
For the unit-scale <Laplace distribution>, integrating separately over the two half-lines gives the <moment-generating function>
$$
M(t)=\frac12\int_0^\infty e^{-(1-t)x}\,dx
+\frac12\int_{-\infty}^0e^{(1+t)x}\,dx
=\frac1{2(1-t)}+\frac1{2(1+t)}
=\frac1{1-t^2},
$$
for $|t|<1$; the appropriate half-line integral diverges outside that interval. Therefore
$$
\boxed{K(t)=-\log(1-t^2),\qquad |t|<1.}
$$
Its derivatives needed for the <saddlepoint density approximation> are
$$
K'(t)=\frac{2t}{1-t^2},\qquad
K''(t)=\frac{2(1+t^2)}{(1-t^2)^2}.
$$
Set $D=\sqrt{1+y^2}$. Solving $K'(\widehat t)=y$ and selecting the root in $(-1,1)$ gives the stable formula
$$
\widehat t=\frac{y}{1+D},
$$
including $\widehat t=0$ at $y=0$. The other quadratic root is outside the domain. At the correct saddlepoint,
$$
1-\widehat t^2=\frac2{1+D},\qquad
\widehat t y=D-1,\qquad
K''(\widehat t)=D(1+D).
$$
Substitution gives the <saddlepoint density of a Laplace sample mean>:
$$
\boxed{\widehat f_{\bar Y}(y)=
\sqrt{\frac{n}{2\pi D(1+D)}}
\exp\!\left[n\left\{\log\frac{1+D}{2}-D+1\right\}\right],
\quad D=\sqrt{1+y^2}.}
$$
It is symmetric in $y$, as the underlying <Laplace distribution> requires. At zero its value is $\sqrt{n/(4\pi)}$, consistent with the local <normal approximation> having <variance> $2/n$. This is an approximation, not an exact finite-$n$ density or an automatically normalized one.
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