Let $x=x_N\ldots x_0$, $y=y_M\ldots y_0$ be integers with base-$b$ digits $0 \le x_i,y_j < b$. Concatenate them via
$$x \otimes_b y\triangleq x_N\ldots x_0 y_M\ldots y_0=b^{len_b(y)}x+y=b^{1-\{\log_b{y}\}}xy+y$$
where $len_b(y)=M=\lfloor\log_b{y}\rfloor+1=\log_b{y}-\{\log_b{y}\}+1$, thus extending to all $b \ne 1$.
$\lim_{b \rightarrow 1} b^{1-\{\log_b{y}\}}=1$ when $\log(y) > 0$, so that $\lim_{b \rightarrow 1}x \otimes_b y = xy+y$.
Q: When $b=1$ (unary), concatenation should correspond to addition. Why is $\lim_{b\rightarrow 1} x\otimes_b y \ne x+y$?
I understand there are some difficulties in viewing unary as an honest base.
(Too long for a comment.)
This works for positional numerals in base $b \ne 1$, and it works because 1) multiplication by $b$ is equivalent to a left-shift of all digits with a $0$ inserted at the rightmost position, and 2) $\text{len}(y)=\lfloor\log_b{y}\rfloor+1$ represents the number of digits in the representation of $y$.
But in the unary numeral system 1) there is no digit $0$ and no notion of left-shifting the digits with "nothing" inserted at the right, and 2) the length of the representation is simply $\text{len}(y)=y$. Because of that, it is not possible to define concatenation for unary numerals as a limit case of concatenation for positional numerals of base $b \ne 1$. Instead, concatenation for unary numerals is simply $\,x \otimes_1 y = x + y\,$.