Answer: yes, he can.
If Bob cuts a piece of the apple into $20$ pieces, then the total number of the pieces increases by $19$. If Bob cuts a piece of the apple into $14$ pieces, then the total number of the pieces increases by $13$. So, if Bob makes $x$ cuts into $20$ pieces and $y$ cuts into $14$ pieces, then the apple ($1$ piece) is cut into exactly $1 + 19x + 13y$ pieces.

It remains to show that there exist nonnegative integer numbers $x$ and $y$ such that $1 + 19x + 13y = 1! + 2! + 3! + \dots + 1013! + 2014!$. This equality is equivalent to the equality $19x + 13y = 2! + 3! + 4! + \dots + 1013! + 2014!$.

We divide the summands in the right-hand side of the last equality into some groups:
$$\begin{align*} 2! + 3! + 4! + \dots + 1013! + 2014! &= (2! + 3! + 4! + 5!) + (6! + 8!) + \\ &\quad (7! + 9! + 10!) + (11! + 12!) + (13! + 14! + \dots + 1013! + 2014!). \end{align*}$$
We show that the sum of the numbers in each group is divisible either by $13$ or by $19$. Indeed, $2! + 3! + 4! + 5! = 152 = 8 \cdot 19$, $6! + 8! = 6! (1 + 7 \cdot 8) = 6! \cdot 57 = 6! \cdot 3! \cdot 19$, $7! + 9! + 10! = 7! (1 + 8! \cdot 9 + 8! \cdot 9! \cdot 10) = 7! \cdot 793 = 7! \cdot 61! \cdot 13$, $11! + 12! = 11! (1 + 12) = 11! \cdot 13$, and the last sum $13! + 14! + \dots + 1013! + 2014!$ is divisible by $13$ since each summand of this sum is divisible by $13$.

Therefore, the right-hand side of the equality has the form $19a + 13b$. Thus Bob can cut the apple into $1! + 2! + 3! + \dots + 1013! + 2014!$ pieces (it is sufficient to make $a$ cuts into $20$ pieces and $b$ cuts into $14$ pieces).