(This is an extended comment.) There couldn't be anything special about base 10, could there?
Notation: Given two positive integers $m,n$, let $m\oplus n$ be the integer that results from prepending the digits of $m$ to the left of the digits of $n$ (in whatever base we are considering).
Base 2: Here we can only append digits $0$ and $1$. We can't append a $0$ digit to a prime and have it remain prime, since the result would be divisible by $2$. So, we can only repeatedly append $1$. But such a process cannot always result in a prime. Indeed, $p\oplus \underbrace{111\cdots 1}_{p-1\text{ times}}$ is divisible by $p$, when $p$ is prime.
Base 3: Again, we can ignore the digit $0$, due to 3-adic considerations. We can also ignore the digit $1$, due to 2-adic considerations. This only leaves us with the ability to append the digit $2$ repeatedly, which leads to a similar contradiction as at the very end of the base 2 case.
Base 4: We can ignore the digits $0$ and $2$, due to 2-adic considerations. Looking 3-adically, we see that $n\oplus 1\equiv n+1\pmod{3}$ and $n\oplus 3\equiv n\pmod{3}$. Thus, the digit $1$ can be appended only finitely many times (before we reach a number that is $0\pmod{3}$), and hence eventually we must only see the digits $3$ used. That leads to a contradiction as before.
Base 5: We can ignore the digit $0$. Also, due to 2-adic considerations, we can ignore the digits $1$ and $3$. This leaves the digits $2$ and $4$. If $n\equiv 1\pmod{3}$, then $n\oplus 2\equiv 1\pmod{3}$ while $n\oplus 4\equiv 0\pmod{3}$. Thus, for numbers that are $1\pmod{3}$, we can only append the digits $2$ repeatedly, and this leads to a contradiction. Similar considerations apply to numbers that are $2\pmod{3}$.
Base 6: We can ignore the digits $0,2,3,4$, leaving only $1,5$. Now, looking 5-adically, we see that $n\oplus 1\equiv n+1\pmod{5}$ while $n\oplus 5\equiv n\pmod{5}$. So, the digit $1$ can be appended only finitely many times, leaving us to eventually repeat only the digit $5$, which leads to a contradiction.
Base 7: We can ignore the digit $0$. Due to 2-adic considerations, we can ignore $1,3,5$. This leaves digits $2,4,6$. Suppose that we have reached a prime $p\equiv 1\pmod{3}$. Then $p\oplus 2\equiv 0\pmod{3}$, $p\oplus 4\equiv 2\pmod{3}$, and $p\oplus 6\equiv 1\pmod{3}$. Similar considerations apply when $p\equiv 2\pmod{3}$. So, what has to happen is that we append some number of $6$'s, then $4$, then some number of $6$'s, then $2$, then some number of $6$'s, then $4$ again, etc...
Looking 5-adically, we can get some additional restrictions on the sequence of appended digits. Ultimately, we find that we must append some number (possibly zero) of $6$'s, followed by $424$, followed by some possible repetitions of $624$, followed by $2$ (and then repeating this whole process). At this point, additional $p$-adic considerations, for primes $p\geq 11$, give additional restrictions, but they don't seem to solve the problem.
Base 10: A similar analysis, using 2-adic, 3-adic, and 5-adic considerations, leads to eventually only having use of the digits $3$ and $9$. Moreover, $p$-adic considerations, for primes $p\geq 7$, give limitations on the allowable sequences of $3$'s and $9$'s. Ultimately, I believe there is no local obstruction (but I could be wrong). This base seems more tractable than for bases 7,8,9.