Based on Powers

The factors of $19607$ are $1$ , $7$ , $2801$ , and $19607$ , where $7$ and $2801$ are primes. It is obvious that $x \ne 1$ and $x \ne 19607$ and since $2801^2 > 19607$ and $2801^1+2801^1+2801^1+2801^1+2801^1 < 19607$ , $x \ne 2801$ , thus we must have $x=7$ .

Assuming that $a \le b \le c \le d \le e$ and that $a=1$ so that:

$\begin{aligned} x^a + x^b + x^c + x^d + x^e & = 7\left(1+ 7^{b-1} + 7^{c-1} + 7^{d-1} + 7^{e-1}\right) \\ & = 7(2801) \end{aligned}$

Assuming that $1+ 7^{b-1} + 7^{c-1} + 7^{d-1} + 7^{e-1}$ is a geometric progression sum with common ratio $7$ , then:

$\begin{aligned} \sum_{k=0}^4 7^k & = \frac{7^5-1}{7-1} = \frac{16806}{6} = 2801 \end{aligned}$ , which is true.

Then we have $a=1$ , $b=2$ , $c=3$ , $d=4$ , $e=5$ and $x=1$ and that $a+b+c+d+e+x$ $= 1+2+3+4+5+7 = \boxed{22}$ .