I was wrong

@above Minivan can start with any prime, then he adds the digits. For example: $7\to 73 \to 733 \to 7333 \to 73331$ (he cannot add $1$ or $3$ now)

Here is the real solution. Let the starting prime $p$ and we assume it is possible. Then divide into cases : Case 1. $p = 2$ or $5$. Since $p \equiv 2 (mod 3)$, $1$ cannot be used. And since $233333 = 353 \times 661$ and $533 = 13 \times 41$, it is done. Case 2. $p = 3$. We do a bit of calculation. $3111 = 3 \times 17 \times 61$, $3113 = 11 \times 283$. $3131 = 31 \times 101$, $3133 = 13 \times 241$, $33 = 3 \times 11$. Case 3. $\gcd(p,30)=1$. By $ \mod 3$, $1$ can be used at most twice. So there exists a moment when only $3$ is being used. Let the number $A$. (so it does $p \to \cdots \to A \to A3 \to A33 \to \cdots$.) Consider a prime $q$ divding $A$. Since $\gcd(p,30)=1$, $\gcd(q,30)=1$ works too. We see the number $A33 \cdots 3$ where $3$ is used $q-1$ times. Since $q \mid 10^{q-1}-1 = 99 \cdots 9$ and $\gcd(q,3)=1$, $q \mid A33 \cdots 3$, as desired.

Solved with GoodMorning. No. Suppose he could. Due to mod $3$ reasons, he can't add $1$s infinitely many times, hence at some point he will only be adding $3$s. Let $N$ be the number before this happens. We find that $N \cdot 10^k + \frac{10^k - 1}{3} = f(k)$ is a prime for each positive integer $k$. Notice that $f(k)$ is a prime, and we see by FLT that $f(k) \equiv f(k + f(k) - 1) \pmod{f(k)}$ (since $f(k)\nmid 10$), meaning that $f(k) \mid f(k + f(k) - 1)$. But $f(k + f(k) - 1) > f(k)$, so it isn't prime, contradiction.