Actually, part (a) is also easy to prove without the assumption $e_0=1$ (although the argument in #2 does not quite work then). But what about part (b)? Is it also true for general (positive) initial values?

To address Tintarn: I think one can show (b) for any general initial condition. Here is my argument. Note that $e_n-a=e_{n-1}(e_{n-1}-a)$, thus $e_n=e_{n-1}^2-ae_{n-1}+a$ for $n\ge 2$. Now, if a prime $p$ and $n$ are such that $p\mid e_n$ ($n\ge 2$) then $p\mid x^2-ax+a$ for a suitable $x$, namely $p\mid (2x-a)^2-a^2+4a$. That is, $a^2-4a$ is a quadratic residue, modulo $p$. Now, unless $a=4$, $a^2-4a$ is not a square, and in this case there exists a prime $q$ (in fact infinitely many) for which $a^2-4a$ is a non-residue, thus for any such prime $q$, $q$ cannot divide any term of the sequence provided $q\nmid e_0$ and $q\nmid e_1=a+1$. For $a=4$, we have $e_n = (e_{n-1}-2)^2$ for $n\ge 2$. If $p\mid e_n$ for a suitable $n$ then $e_{n-1}\equiv 2\pmod{p}$, but as $e_{n-1}$ is a square as well, it follows $2$ is a quadratic residue modulo $p$. There are infinitely many primes $p$ for which $2$ is quadratic non-residue (take $p\equiv \pm 3\pmod{8}$), which also shows no matter what the initialization is, (b) always holds.

I have just realised that I forgot to post my solution for part (a).

Part a) Suppose only finetely many primes devides some element of the set $A=\{e_n|n\ge 0\}$.Then only finitelty many primes devide some of the element of set $B=\{a_n| a_n=e_0e_1\dots e_n\}$.By Kobayashi's theorem there are infinitely many primes devide some of the elements of $B+a=\{a_n+a|a_n=e_0e_1\dots e_n\}=B-\{1\}$.But it is a contradiction of our initial assumption that the set $B$ has only finitely many primes dividing some elements of $B$. Hence we are done.$\blacksquare$ For part (b)If $a\neq 1$ then as $e_n\equiv 1\pmod {a}$ Taking a prime $p$ deviding $a$ we get $gcd(e_n,p)= gcd(1,p)=1$. If $a=1$ then as $e_n=1+e_0e_1\dots e_{n-1}=1+e_{n-1}(e_{n-1}-1)$ hence $5$ doesn't devide any of $e_n$ as 5 does not devide $x^2-x+1$.$\blacksquare$

Claim: All terms of the sequence are relatively prime to $a$. Proof: This is by induction with base case $0$. If it is true for everything up to $n - 1$, then $\prod_{k=0}^{n-1} e_k$ is relatively prime to $a$, so $e_n = a + \prod_{k=0}^{n-1} e_k$ must also be relatively prime to $a$. a) If not, then some prime number must divide infinitely many elements of the sequence. Suppose it's $p$ and it divides $e_x$ and $e_y$ for some $x < y$. Notice that $e_y = a + \text{ a multiple of } e_x$, so $e_y \equiv a\pmod{e_x}$, implying that $p\mid a$, contradiction since $\gcd(e_x, a) = 1$. b) If $a\ne 1$, just take any prime divisor of $a$. Now assume $a = 1$. We claim that $5$ doesn't divide any element of the sequence. Notice that $e_n = e_{n-1}^2 - e_{n-1} + 1$, but $5$ can't divide this.