I set out intending to post this as a comment, but perhaps it sits better here. The method that follows suggests a general formula to produce such relationships from second order linear recurrences. Modulo issues of convergence, it's more or less elementary. Some text is required to set up the machinery, but the substance of the argument is really just a few lines.
Context
Let $\Lambda$ be the complex vector space of functions $f\colon \mathbb{Z}\to \mathbb{C}$ for which there exists a constant $K_{f}\in\mathbb{R}_{\geq 1}$ such that $$\lim_{n^{-1}\to 0}\left(\frac{f\left(n\right)}{K_{f}^{\left|n\right|}}\right)\ =\ 0.$$ (The addition of $\Lambda$ and $\mathbb{C}$-action on $\Lambda$ are computed pointwise.) Now let $$S\ \colon\ \Lambda \to \Lambda,\ \ Sf\left(n\right)\ :=\ f\left(n+1\right)$$ be the ($\mathbb{C}$-linear) shift operator. Define $S^{-1}$ analogously, it being the (unique) inverse of $S$. For any polynomial $P\in\mathbb{C}\left[S,S^{-1}\right]$, we have a well-defined $\mathbb{C}$-linear map $$P\ \colon\ \Lambda\to\Lambda$$ (perhaps implicitly using that $S$ and $S^{-1}$ commute; the multiplicative structure of the ring corresponds to the composition of operators so that in particular $1\in \mathbb{C}\left[S,S^{-1}\right]$ is identified with the identity). For such $P$, we consider in turn the $\mathbb{C}$-linear map $$e^{P}\ :=\ \sum_{d\in\mathbb{N}} \frac{P^{d}}{d!}\ \colon\ \Lambda \to \Lambda.$$ The well-definition of $e^{P}$ consists of its pointwise absolute convergence and its taking functions in $\Lambda$ to functions in $\Lambda$; both follow from standard arguments$^{\text{1}}$. Also by a standard argument$^{\text{2}}$, $P_{0}$ commutes not only with $P_{1}$ but moreover with $e^{P_{1}}$ for any $P_{0},P_{1}\in \mathbb{C}\left[S,S^{-1}\right]$. By yet another standard argument$^{\text{3}}$, $$e^{P_{0}}e^{P_{1}}\ =\ e^{P_{0}+P_{1}}\ =\ e^{P_{1}}e^{P_{0}}.$$ Finally, it's clear that $e^{1}$ is the operator that scales inputs by $e$ and that if for some $f\in\Lambda$ and $P\in \mathbb{C}\left[S,S^{-1}\right]$ we have that $Pf = 0$, then $\left(e^{P}-1\right)f\ =\ 0$.
The argument
Now fix $f$ the Fibonacci sequence; this is in $Λ$ as by an elementary induction we have that $\left|f\left(n\right)\right|\leq 2^{\left|n\right|}$. The defining linear recurrence of $f$ is that $$\left(S-S^{-1}-1\right)f\ =\ 0.$$ From this it follows that for each $n\in\mathbb{Z}$, \begin{align*}\sum_{d\in\mathbb{N}}\frac{f\left(n+d\right)}{d!}\ :&=\ e^{S}f\left(n\right)\\ &=\ e^{S^{-1}+1}e^{S-S^{-1}-1}f\left(n\right)\\ &=\ e^{S^{-1}+1}f\left(n\right)\\ &=\ e^{1} e^{S^{-1}}f\left(n\right)\\ :&=\ e\sum_{d\in\mathbb{N}}\frac{f\left(n-d\right)}{d!},\end{align*} and modulo the question of the vanishing of the denominator in the expression in the OP (which is contingent on not only the recurrence, but also the specific initial conditions of the sequence), this is precisely what was to be demonstrated. $\blacksquare$
Notes
All instances of "convergence" in reference sums in what follows refer specifically to absolute convergence pointwise in $n$.
For any $f\in \Lambda$ with $K_{f}\in\mathbb{R}_{\geq 1}$ such that $\lim_{n^{-1}\to 0}\left(f\left(n\right)K_{f}^{-\left|n\right|}\right)\ =\ 0$ and any $P\in \mathbb{C}\left[S,S^{-1}\right]$ whose terms have sum of norms of coefficients $L\in\mathbb{R}_{\geq 0}$ and maximum absolute degree $h\in\mathbb{N}$, there must (by the limit hypothesis) for every $\varepsilon > 0$ exist an $M_{\varepsilon}\in\mathbb{R}_{\geq 0}$ such that $\left|f\left(n\right)\right|\leq \max\left(M_{\varepsilon},\varepsilon K_{f}^{\left|n\right|}\right)$. It's then not hard to see (by expanding $P^{d}$) that $$\left|\frac{P^{d}}{d!}f\left(n\right)\right|\ \leq\ M_{\varepsilon}\frac{L^{d}}{d!}+\varepsilon\frac{\left(LK_{f}^{h}\right)^{d}}{d!}K_{f}^{\left|n\right|}$$ and thus that $$\left|e^{P}f\left(n\right)\right|\ \leq\ M_{\varepsilon}e^{L}+\varepsilon e^{LK_{f}^{h}}K_{f}^{\left|n\right|}$$ (so that the sum on the left necessarily converges). As the choice of $\varepsilon$ was arbitrary, it follows that $$\lim_{n^{-1}\to 0}\left(\frac{e^{P}f\left(n\right)}{K_{f}^{\left|n\right|}}\right)\ =\ 0,$$ completing the argument for both claims.
As $P_{0}$ plainly commutes with the partial sums that converge to $e^{P_{1}}$, it must commute with $e^{P_{1}}$ itself.
Expand the sums on the left and right as is justified by convergence and then use commutativity to collect and rearrange the terms to obtain the sum of binomial expansions implied by the middle.