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.

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 $> 1$

I will use $*$ for your concatenation operation, and $\cdot$ for true multiplication.

1) If you allow for leading zeroes to be ingored, then $0*0=00=0$. Notice that for all pairs of non-zero natural numbers, $a\cdot b\leq a\cdot10^{\lceil\log_{10}(b)\rceil}$ but that $a*b>a\cdot10^{\lceil\log_{10}(b)\rceil}$. If $a=0$ and $b\neq 0$ and we ignore leading zeroes, then we are still an overestimation, and if $a\neq0$ and $b=0$ then we are still an overstimation.

2) Based on my crude estimates above, yes, there is a way of putting bounds on the size of the overestimation, but I don't know if you can do much better honestly.

3) Also based on my crude estimations, if you replace the logarithms with different bases I'm fairly sure that this shows no base greater than 1 works. Base 1 itself actually has both types of behaviour; $1*11=111>11$, but $111*111=111111<111111111$. Additionally, you have an example of equality in $11*11=1111$. There is of course the issue that $0$ is a strange object to try and work with in base 1, so let's just ignore that for now...

I can't muster the strength of will to try and prove anything for negative bases. I suspect that negative bases less than $-1$ will fail, but it is easy to see that in base $-1$ there are trivial representations that will also give you equality; $11*11=1111$ where everything in sight is $0$ in base 10.

So it was impossible with natural numbers (see other answers) But if you willing to bend the rules a bit, by making that when decimal numbers are involved $A.a \times B.b = AB.ab$

then you can find

$$x.99999999999... \times 9.9999999999... = x9.99999999999...$$

or more compactly

$$x.(9) \times 9.(9) = x9.(9)$$

this is possible because $x.(9) = x + 1$

and for an example if $x=4$

$5 \times 10 = 50 \Leftrightarrow 4.(9) \times 9.(9) = 49.(9)$