Are there any two numbers such that multiplying them together is the same as putting their digits next to each other?

I have two natural numbers, $A$ and $B$, such that $A \times B = AB$.

Do any such numbers exist? For example, if $20$ and $18$ were such numbers then $20 \times 18 = 2018$.

Lets put aside the trivial answer $A=0$ and $B=0$ and consider both $A, B>0$.

You want numbers such that $A\times B = A\times10^k + B$ where $k$ is the number of digits of $B$, that is with $10^{k-1}\leq B < 10^k$. So you need $B=10^k+\dfrac{B}{A}$ with $B<10^k$. From which results $B>10^k$ and $B<10^k$.

So if there's no mistake there, the answer is no.

There is the pathological example $A=B=0$.

For the rest:

Let $B$ have $m$ digits. We have $AB= A*10^m+B$ We want $AB=A* B$,

We have $$A*10^m+B-A*B = A*(10^m-B)+B>B,$$ because $10^m>B$ as $B$ has $m$ digits. So you always overestimate by at least $B$.

From this its also clear, that the result is independent of the chosen base.

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Let me generealise a bit:
If we allow $A$ to be negative, we need to change the condition to $AB=A*10^m-B$.
This leads to $$A=\frac B {10^m-B}, $$ but then the right hand side is positive and again we get a contradiction.

So the last possibility is $B<0$. But then we first need to define $AB$.
A natural way to do this would be $A(-B)$.
Then we have for $A>0$: $AB=A*10^m-B$ which implies $$A=\frac B {10^m-B}. $$ This time, there is no contradiction because of sign issues.
However, now $-B>0$, hence $|10^m-B|>|B|$, so the right hand side is no integer.

The analogous argument gives also a contradiction for $A,B>0$.

So even in the more generalised setting, the answer is no.

If $B$ has $n$ digits then $10^{n-1} \le B <10^n$ and we want $AB = 10^nA + B$ or

But $B<10^n$ so $AB < 10^nA \le 10^nA + B$

So 1) Yes over compensation always

2) by $10^nA + B - AB = 10^{[\log_{10}(B)]+1}A + B - AB$

3) The same argument applies to any base $> 1$