A variant of my previous answer. It is commonly believed that all irrational algebraic numbers are normal. If this is the case, then there can be at most two algebraic numbers (up to shifts by rationals) that can be obtained by transposing digits of $e$ and $\pi$.
To see this, suppose that there are three algebraic numbers $\alpha,\beta,\gamma$, no two of which differ by a rational, that can all be attained by transposing digits of $e$ and $\pi$. By the pigeonhole principle, we see that for each natural number $k$, at least one of the pairs $(\alpha,\beta)$, $(\alpha,\gamma)$, $(\beta,\gamma)$ agree at the $k^{th}$ digit. By the pigeonhole principle again, this means that one of these pairs agrees on a set of digits of (upper) density at least $1/3$. Without loss of generality we can assume that the pair $(\alpha,\beta)$ has this property, and that $\beta > \alpha$. But then, by long subtraction, the difference $\beta - \alpha$ will have digits $0$ or $9$ on a set of digits of upper density at least $1/3$, which contradicts the normality of $\beta-\alpha$. (Now I need the base to be at least seven!)
It might be possible to upgrade "up to shifts by rationals" in the above claim by "up to shifts by terminating decimals", but I have not strenuously attempted to do this.