The identity given is simpler than it seems. Suppose
that $z$ is any nonzero complex number. Then

$$ \sum_{k=0}^\infty \frac{z^{n+k}}{k!} =
z^n \sum_{k=0}^\infty \frac{z^k}{k!} = e^z z^n, \quad
\sum_{k=0}^\infty \frac{z^{n-k}}{k!} =
z^n \sum_{k=0}^\infty \frac{z^{-k}}{k!} = e^{1/z} z^n. \tag1 $$

Now, if $\,a_n := u\,\alpha^n + v\,\beta ^n\,$ for any nonzero
$\,\alpha,\beta\,$ then

$$ \sum_{k=0}^\infty \frac{a_{n+k}}{k!} =
u\, e^\alpha \alpha^n + v\, e^\beta \beta^n, \quad
\sum_{k=0}^\infty \frac{a_{n-k}}{k!} =
u\, e^{1/\alpha} \alpha^n + v\, e^{1/\beta} \beta^n. \tag2 $$

Divide the two summations to get

$$ \frac{ \sum_{k=0}^\infty \frac{a_{n+k}}{k!} }
{ \sum_{k=0}^\infty \frac{a_{n-k}}{k!} } =
\frac{ u\, e^\alpha \alpha^n + v\, e^\beta \beta^n }
{ u\, e^{1/\alpha} \alpha^n + v\, e^{1/\beta} \beta^n}. \tag3 $$

For the Fibonacci sequence, $\,\alpha-\frac1\alpha =
\beta -\frac1\beta = 1.$ The right side thus simplifies to $\,e^1 = e.$

In general, if $\,\alpha\beta=-1,\,$ then
$\, \gamma := \alpha-\frac1\alpha = \beta -\frac1\beta,\,$
and the right side simplifies to $\,e^\gamma.$

There is an equivalent approach to a proof. Let
$\,z\,$ be either $\,\alpha\,$ or $\,\beta.\,$
Then $\, z = \gamma + 1/z\,$ and

$$ z^{n+k} = z^n(\gamma+1/z)^k = \sum_{j=0}^k {k\choose j}
\gamma^{k-j}z^{n-j}. \tag4 $$

Appy this to the sequence $\,a_n\,$ to get

$$ a_{n+k} = \sum_{j=0}^k {k\choose j}
\gamma^{k-j} a_{n-j}. \tag5 $$

Now sum to get

$$ \sum_{k=0}^{\infty} \frac{a_{n+k}}{k!} = \sum_{k=0}^{\infty}
\frac{ \sum_{j=0}^k {k \choose j} \gamma^{k-j} a_{n-j}}{k!} = \sum_{k=0}^{\infty}\sum_{j=0}^{k}\frac{\gamma^{k-j}}{j!(k-j)!} a_{n-j}\\
= \sum_{j=0}^{\infty} \sum_{k-j=0}^{\infty} \frac{\gamma^{k-j}}{(k-j)!}
\frac{a_{n-j}}{j!} = \sum_{j=0}^{\infty} e^\gamma \frac{a_{n-j}}{j!} =
e^\gamma \sum_{j=0}^{\infty} \frac{a_{n-j}}{j!}. \tag6 $$

This is another proof of the general result of equation $(3)$.