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 for sake of contradiction 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. It's also worth noting that this is an example of an ineffective argument, in that no bound whatsoever is provided on the height of the algebraic numbers that might still be obtainable by transposing digits of $e$ and $\pi$, even if one had some quantitative normality bound on algebraic numbers depending on the height.

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 for sake of contradiction 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. It's also worth noting that this is an example of an ineffective argument, in that no bound whatsoever is provided on the height of the algebraic numbers that might still be obtainable by transposing digits of $e$ and $\pi$, even if one had some quantitative normality bound on algebraic numbers depending on the height.

p.s. We can combine the two answers: if we assume that the sum of $\pi$ and an algebraic number, or the sum of $e$ and an algebraic number, is always normal, then the answer to the original question is positive: every transposition of $\pi$ and $e$ is transcendental. For if there was an algebraic number $\alpha$ that was achievable as a transposition, then it would have to share at least half its digits with $\pi$ or $e$, and hence one of $|\pi-\alpha|$ or $|e-\alpha|$ would have digits $0$ or $9$ on a set of upper density at least $1/2$, contradicting the normality of these numbers. (Now I need base at least five.)

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.

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 for sake of contradiction 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. It's also worth noting that this is an example of an ineffective argument, in that no bound whatsoever is provided on the height of the algebraic numbers that might still be obtainable by transposing digits of $e$ and $\pi$, even if one had some quantitative normality bound on algebraic numbers depending on the height.