This is not possible. In Kuipers, L. and Niederreiter, H. Uniform Distribution of Sequences (1974), you can find the following exercise, on page $77$:
If $b_1$ and $b_2$ are integers $≥2$, such that one is a rational power of the other, then $a$ is normal to the base $b_1$ if and only if $a$ is normal to the base $b_2$.
(This is also theorem 2.7. in Bailey, D. H. and Crandall, R. E. On the Random Character of Fundamental Constant Expansions, Experiment. Math., Volume 10, Issue 2 (2001), 175-190.)
Therefore, if $x$ is a real number that is normal in every base but one, say $b_1$, then actually it wouldn't be normal to the base $b_1^2 \neq b_1$, contradicting the assumption. The situation is similar for numbers that are non-normal in every base except one.
We may represent integer $\ n>0\ $ as $\ n\ =\ b\cdot 10^d+M\cdot 10 + a\ $
(so that $\ a>0\ $ or $\ M>0\ $ or else we would have $ a=M=b=0)\ $
so that
$$ 2\cdot(a\!\cdot\! 10^d + M\!\cdot\! 10 + b)
\ =\ b\!\cdot 10^d+M\!\cdot 10 + a $$
where $\ a\ b\ $ are decimal digits (i.e. $\ 0\ldots 9),\ $ and $\ M\ $ is
a non-negative integer such that $\ M<10^{d-1}.$ Equivalently,
$$ (2\!\cdot\! 10^d-1)\!\cdot a + 10\!\cdot\!M
\,\ =\,\ (10^d-2)\!\cdot\! b $$
or
$$ b\,\ =\,\ 2\!\cdot\!a\ +\ \frac{10\!\cdot\!M\ +\ 3\!\cdot\!a}{10^d-2} $$
Thus,
$$ 0\ <\ b-2\cdot a\ = \frac{10\!\cdot\!M\ +\ 3\!\cdot\!a}{10^d-2}\
\le\ 1\ +\ \frac{3\cdot a-8}{10^d-2} $$
Also if we had $\ d\ge 2\ $ then this very last fraction has an
absolute value $\ < 1\ $ so that
$$ \frac{10\!\cdot\!M\ +\ 3\!\cdot\!a}{10^d-2}\ =\ 1 $$
while, knowing that $\ a<5\ $ (of course!),
$$ 10\!\cdot\!M\ +\ 3\!\cdot\!a\,\ \not\equiv
\,\ 10^d-2\quad \mod 10 $$
This contradiction shows that $\ d\le 1.\ $ Thus there are
only 2-digit solutions (if any).
In particular, $\ M=0.\ $ This simplifies the equation:
$$ 19\cdot a\ \ =\ \ 8\cdot b $$
However, $19$ does not divide $\ 8\cdot b.\ $ Thus, after all,
$\qquad\qquad$ THERE ARE NO SOLUTIONS.
Best Answer
For bases sufficiently large (as @Goos noted, there are examples for low bases), it is impossible. I will use the facts that there must be as many numbers as the base, and that the digits must sum to the base (since they count the digits in the number).
Take the base $=n$, and consider the first digit $k$. $k$ cannot be zero, because that would say there are zero zero's, which there wouldn't be! So there are $k$ zero's, and at least one $k$ (for now), meaning we have $$n-k-1$$ digits left to fill. These digits must add to the remaining $n-k-k\cdot 0=n-k$, and the only $n-k-1$ digits which sum to $n-k$ are $$1,1,\ldots,1,2$$ (we have $n-k-2$ one's). I will eliminate $k=1,2$ later. So we have $2$ of something, not $k$'s and not zero's. We must have $2$ one's. And then we have one $k$ and one $2$, which works. The condition that the numbers sum to the base gives $$k+1+1+2+k\cdot 0=k+4=n \implies k=n-4$$ so the only possible self-descriptive number is $$(n-4)(2)(1)0\ldots(1)0\ldots0.$$
If $k=1$, then there will have to be two one's, and one two, and one zero. This gives the only possibility $$(1210)_4$$ in base $4$.
If $k=2$, then there must be two two's. We have two zero's, so there are one or no one's. The two possibilities here are $$(21200)_5,\quad (2020)_4$$ So we have enumerated all self-descriptive numbers.