So I was thinking say you have a linear differential operator such as the exponential differential one which is renown in some fields in physics: $$e^{\mathrm D_x}\equiv\sum_{n=0}^\infty\frac{\mathrm D_x^n}{n!},\text{ where } \mathrm D_x^n\equiv\frac{\mathrm d^n}{\mathrm dx^n},$$ and you applied it to a product of functions $u(x)\cdot v(x)$.
Then what would the "product rule" be for this operator? I.e., $$e^{\mathrm D_x}[u\cdot v]=?$$
APPROACH: Is this it? $$\sum_{n=0}^\infty\frac{1}{n!}\frac{\mathrm d^n}{\mathrm dx^n}(u\cdot v)=\sum_{n=0}^\infty\frac{1}{n!}\sum_{i=0}^n{n\choose i}u^{(n-i)}v^{(i)}=\sum_{n=0}^\infty\sum_{i=0}^n\frac{u^{(n-i)}}{(n-i)!}\frac{v^{(i)}}{i!}\overset{j=n-i}{=}\sum_{i=0}^\infty\sum_{j=0}^\infty\frac{u^{(j)}}{j!}\frac{v^{(i)}}{i!}$$
Your approach is correct. You can go on by introducing an auxiliary variable $t$ in order to make the respective Taylor series of $u$ and $v$ appear, as it follows : $$ \begin{align} e^{D_x}(uv)(x) &= \sum_{i,j\ge0} \frac{u^{(i)}(x)}{i!}\frac{v^{(j)}(x)}{j!} \\ &= \left[\sum_{i=0}^\infty \frac{u^{(i)}(x)}{i!}t^i \sum_{j=0}^\infty \frac{v^{(j)}(x)}{j!}t^j\right]_{t=1} \\ &= \left[u(x+t)v(x+t)\right]_{t=1} \\ &= u(x+1)v(x+1) \end{align} $$ This result is not surprising, since the operator $e^{D_x}$ is nothing else than the translation operator.