Fibonacci series captures Euler $e=2.718\dots$ The Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$ allows values for all indices $n\in\mathbb{Z}$. There is an almost endless list of properties of these numbers in all sorts of ways. The below question might even be known. Yet, if true, I like to ask for alternative proofs.

Question. Does the following identity hold? We have
  $$\frac{\sum_{k=0}^{\infty}\frac{F_{n+k}}{k!}}{\sum_{k=0}^{\infty}\frac{F_{n-k}}{k!}}=e,$$
  independent of $n\in\mathbb{Z}$.

 A: It follows from the identity
$$F_{n+k} =  \sum_{j=0}^k {k \choose j} F_{n-j}$$ 
which is obtained by applying the standard recurrence $k$ times to the left side, each time splitting up each term into two terms.
Indeed, this gives
$$\sum_{k=0}^{\infty} \frac{F_{n+k}}{k!} = \sum_{k=0}^{\infty} \frac{ \sum_{j=0}^k {k \choose j} F_{n-j}}{k!}= \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{1}{j! (k-j)!} F_{n-j} = \sum_{j=0}^{\infty} \sum_{k-j=0}^{\infty} \frac{1}{(k-j)!} \frac{F_{n-j}}{j!} = \sum_{j=0}^{\infty} e \frac{F_{n-j}}{j!}$$
A: Mathematica tells me it's a consequence of the two series (distinguished by $\pm$):
$$\sum_{k=0}^\infty\frac{F_{n\pm k}}{k!}=\frac{e^{\sqrt{5}} \phi^n-(1-\phi)^n}{\sqrt{5}\, \exp(\phi^{\mp 1})},$$
with $\phi$ the golden ratio (the positive solution of $1+1/\phi=\phi$). So the ratio of the $\pm$ series equals $e^{\phi-1/\phi}=e$, and the entire $n$-dependence drops out of the ratio. 
The identity also holds for $n\in\mathbb{R}$, with the usual generalisation $F_z=(\phi^z-\phi^{-z}\cos\pi z)/\sqrt{5}$ of the Fibonacci numbers to real argument $z$.
A: More generally,
$$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$
which is equivalent to Will Sawin's identity.
Similarly,
$$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},\tag{2}$$
and more generally, from the formula
$$(-1)^q F_{p-q}+F_{\!q}\,\phi^p = F_{\!p}\,\phi^q,$$
where $\phi = (1+\sqrt5)/2$, and the analogous formula with $\phi$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$,
$$
e^{(-1)^q F_{p-q}\ x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} 
  = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}.
$$
The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.
Related identities (also involving Lucas numbers) can be found in 
L. Carlitz and H. H. Ferns,
Some Fibonacci and Lucas Identities,
Fibonacci Quarterly 8 (1970), 61–73.
