You have tried to solve it more algebraically than normal people would do. (I usually do the same thing - too scientific an approach - which occasionally makes us less efficient in similar puzzles.)
Spoilers are found below. Please stop reading if you want to solve it yourself.
Each of the two 2-digit expressions are smaller than $b^2$, so their sum is smaller than $2b^2$. Because this is a 3-digit expression, its first digit has to be $1$. So the square is equal to 1. So we have
$$(1@) + (1@) = (1\bigstar 1)$$
Note that it ends with a $1$ - odd digit - even though the left-hand side is even - twice $(1@)$. It can only happen if the base $b$ is odd.
At any rate, the equation simplifies to
$$2(b+@) = b^2+1 + b\bigstar$$
Moving everything non-negative to the right hand side, we have
$$ 2@ = (b-1)^2 + b \bigstar.$$
The first term is a square of an integer, i.e. $4,16,36,64,\dots $ for $b=3,5,7,9,\dots$. Note that the identity above implies that $2@$ can't be smaller than $(b-1)^2$ but $2@<2b$ and if $b\geq 5$, $2b$ is clearly smaller than $(b-1)^2$. So the only chance to find a solution is $b=3$. Then the equation simplifies to
$$ 2@ = 4+3 \bigstar .$$
The left hand side is even, so the right hand side must also be even. Therefore $\bigstar$ is either $0$ or $2$. For $\bigstar=2$ we would get $@=5$ which is too high so the only other option is $\bigstar=0$ and $@=2$ which works:
$$ (12) + (12) = (101).$$
This trinary equation is translated to decimal base as $5+5=10$. Note that those ETs have 3 fingers in total, so if they have an even number of hands, it follows that different hands have different numbers of fingers, for example 1 left finger and 2 right fingers.