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][1],
Fibonacci Quarterly 8 (1970), 61–73.


  [1]: https://www.fq.math.ca/Scanned/8-1/carlitz.pdf