Some people write February 5$^{th}$ using the md format, as $2.5$ or alternatively as $2/5$.
Note that $2.5 \times \frac25 = 1$.
Is there any other date satisfying $a.b \times \frac{a}{b} = 1$?
What can we say if we do not confine $a$ and $b$ as dates?
I'll interpret the 'general case' as:
Find all pairs of positive integers $(a,b)$ such that $a.b\times\frac{a}{b}=1$ where $a.b$ means a concatenation of $a,b$ with a decimal dot in between.
I'll present an almost-proof of this in base 10.
Firstly, note that the condition is equivalent to $a.b=\frac{b}{a}$. Secondly, note that $a+1>a.b\geq a$ so $a+1>\frac{b}{a}\geq a$, that is, $b=a^2+r$ for some $r<a$. So our equation becomes $a.b=\frac{a^2+r}{a}$ which is the same as $0.b=\frac{r}{a}$, or $\frac{a^2+r}{10^k}=\frac{r}{a}$ for $10^k$ being the smallest power of $10$ (strictly) greater than $a^2+r$, which can be weakened to $10(a^2+r)\geq10^k$.
Therefore we want $a^3=(10^k-a)r$ so $10^k-a|a^3\iff 10^k-a|10^{3k}$. Since $a^2<10^k$ what we really want is for some $10^k-\text{small number}|10^{3k}$. Let $10^k-\text{small number}=2^x5^y$ (since it's a factor of a power of $10$). Then we want to find $x,y,k$ such that $10^k-10^{\frac{k}{2}}\leq2^x5^y\leq10^k$. If $(x,y)$ is a solution with $x,y>0$ then so is $(x-1,y-1)$ so if you do a bit of bounding and stuff*, the only solution with $x\neq1$ is $(1,0), k=1$ (this is kind of easy, if I'm right).
Thanks to Gareth for an idea for the next part, which may or may not work. I don't understand anything well enough to say anything so this part is pure guess and hope:
Otherwise, we want $10^k-10^{\frac{k}{2}}\leq5^y\leq10^k$ and that means that $\frac{\log5}{\log{10}}$ has a 'good' rational approximation. Now Baker's theorem may or may not rule this out as unfeasible, depending on what the theorem actually says. I can't work anything out reasonably at this point so I'll call it a proof and wrap up here.
Anybody solving the above will definitely earn my everlasting respect
Anyway, translating our results back to $a,b$, we find that if the yellow stuff is correct then $(a,b)=(2,5)$ is the only solution.
I'd just like to point out that if we were working in a base $p$ where $p$ is prime, then at the point $p^k-a|p^{3k}$ it's pretty obvious there are no solution where $a$ is small compared to $p^k$, so no solutions exist in prime bases.
I think a similar result holds true in bases of the form $p^x$, in that only finitely many solutions are possible (this shouldn't be too hard to prove after reading the above)
Lastly, in the 'any base' case, I think that there' a chance finitely many solution exist for each base.
*bounding and stuff:
We can write $a.b \cdot a/b=1$ as
$$a+\frac b{10}=\frac ba\implies 10a^2+ab-10b=0$$
From @ffao's comment we need only check for $b=\{1, 2, ..., 12\}$ since there are $12$ months in a year. We also want positive integer solutions for $a$; in other words, we want the discriminant $b^2+400b$ to be an integer. Doing this gives the single pair of solutions
$$(a,b)=(2,5)$$
So May the second is very unique!
$a.b*a/b = b$
$a^2/b+a/10^{d_b} = 1$ ($d_b$ being the number of digits of $b$)
$10^{d_b}a^2+ba = b*10^{d_b}$
For the latter to be true, the left-hand side must be divisible by $b$ and $10^{d_b}$, so $ba$ is divisible by $10^{d_b}$ and $10^{d_b}a^2$ is div. by $b$. Since $b>a^2$, $b$ can't be co-prime with $10^{d_b}$.
$ba = (b-a^2)10^{d_b}$
If only one of $a$ or $b$ has a particular prime factor outside of 2 or 5 (let it be $p$, and the other may or may not have another), $b-a^2$ can't be divided by it, which creates a contradiction.
If $a$ is divisible by $p^x$ and $b$ is divisible by $p^y$ ($2x$ != $y$), the left-hand side is divisible by $p^{x+y}$, but the right-hand side is divisible by $p^{min(y,2x)}$, which is also impossible.
If one of them is odd and the other isn't a multiple of 5, let the product of prime powers save for those of 2 or 5 dividing $a$ be $P$, which is $P^2$ for $b$. Since $a(a+1)>b>a^2$, even the product of powers of 2 and 5 is greater for $b$ than the square of that of $a$, meaning $a = 2^{d_b}*P$ and $b = 5^{d_b}*P^2$, which is impossible for $d_b>1$ (only allowing the trivial answer mentioned in the question itself).
If $a = 2^n 5^m P$ and $b = 2^q 5^r P^2$,
$2^{n+q}*5^{m+r}*P^3 / ([2^{min(q,2n)} * 5^{min(r,2m)}] * X * P^2) = 10^{d_b}$
$X=kP$
Powers of 2: $n+q = min(2n,q) + d_b => d_b = n$ or $q-n$
Powers of 5: $m+r = min(2m,r) + d_b => d_b = m$ or $r-m$
If $q<2n$, then $d_b = n = r-m$ and $r>2m$.
$2^q 5^r P^2 < 2^{r-m} 5^{r-m} => r-m=n>q$.
$2^{r-m+q}*5^{m+r}*P^3 = (2^q 5^r P^2 - 2^{2r-2m} 5^{2m} P^2)*2^{r-m} 5^{r-m}$
$2^{r-m+q}*5^{m+r}*P^3 = (2^q 5^2m P^2) * (5^{r-2m} - 2^{2r-2m-q}) *2^{r-m} 5^{r-m}$
$P = 5^{r-2m} - 2^{2r-2m-q}$
If $q>2n$, then $d_b = q-n = m$ and $r<2m$.
$2^q 5^r P^2 < 2^{q-n} 5^{q-n} => q-n=m>r$.
$2^{n+q}*5^{q-n+r}*P^3 = (2^q 5^r P^2 - 2^{2n} 5^{2q-2n} P^2)*2^{q-n} 5^{q-n}$
$2^{n+q}*5^{q-n+r}*P^3 = (2^{2n} 5^r P^2) * (2^{q-2n} - 5^{2q-2n-r})*2^{q-n} 5^{q-n}$
$P = 2^{q-2n} - 5^{2q-2n-r}$
In the second case, $5^{q-n}>5^r$, which means $2^q>5^r => q>2r$. Combining that with $q-n>r$, we get $r>n$, but since $P = 2^{q-2n} - 5^{2q-2n-r}$, $q-2n > 2q-2n-r=>q>2q-r$, which would be wrong.
In the first case, $2^{r-m}>2^q$, which means $2^q<5^r$.
$2^q 5^r - 2^{2r-2m} 5^{2m} = (2^q*5^{2m})(5^{r-2m} - 2^{r-2m-q}) <= 2^{r-m}* 5^m$
$2^q*5^{2m} <= 2^{r-m}* 5^m$ (impossible)
I’m going to hazard a mathematical answer and say
“No”, there is no other such combination, Not with integers in base 10
Discounting 1 and 10 for a second, the only integer multiples of 10 are 5 and 2. To get the ‘1’ that you need for your answer, the day and the month must be multiples.
Back to 1 and 10.. that only leaves us 2 cases to check
$1.10 \times 1/10 = 0.110$
$10.1 \times 10/1 = 101$
Your “unbounded” section is very broad.. I imagine if we start letting
Imaginary numbers, different bases, different calendars, then there will be other such dates
Is there any other date satisfying $a.b\times\frac{a}{b}=1$?
No, if we assume a single digit $b$.
Consider $a.b\times a=b$. We know $b\gt0$, therefore $b\gt a^2$, so $a\ge3$ is impossible.
Now consider $a.b=\frac{b}{a}$. $a=1$ is easy to exclude as $a.b\lt2$ but $\frac{b}{1}\ge2$.
$a=2$ leaves the cases $b=5,6,7,8,9$, but $b\gt5$ means that $\frac{b}{2}\ge3$.
