Why does $0.75 \times 1.25$ equal $0.93$ and not $1$? If percentage points can equal decimal points, then $1.25$ should equal $125$% when multiplying $1.25$ by any number.
Therefore $125$% of $0.75$ should equal $1$ if $0.75$ is also $75$% of $1$. 
Am I missing something super obvious here or what?!
Please help. Thank you!

As @wythagoras has already written, the reason is that
$$(1-a)(1+a) = 1 - a^2.$$ 
If $a \neq 1$, then $1 - a^2 \neq 1$. So your idea never works except in the most trivial case ($1\times1 =1$).
Another way to understand this is to ask whether $$\frac{1}{1-a} \overset{?}{=} 1+a$$
In your example, this would mean to ask whether $$\frac{1}{0.75} \overset{?}{=} 1.25.$$
In fact, for small $a$ (and consequently, even smaller $a^2$), this is a fairly good approximation based on Bernoulli's inequality (set $r=-1$ and $x=-a$ in $(1+x)^r \approx 1+rx$ to see that). But in general, it is not true.
To see that, you can calculate the Taylor expansion of $\frac{1}{1-a}$, you find $$\frac{1}{1-a} = 1 + a \color{green}{+ a^2 + a^3 + a^4 + a^5 + ...}$$
The green part is what you were missing.
So when you want to "compensate" $0.75$ (for $a = 0.25$), you could compute $1 + 0.25 \color{green}{+ 0.25^2 + 0.25^3 + 0.25^4 + 0.25^5 + ...} = 1.333 > 1.25$.
It is easy to see that this always work:
$$(1-a)(1 + a + a^2 + a^3 + a^4 + a^5 + ...) \\
= (1 + a + a^2 + a^3 + a^4 + a^5 + ...) \phantom{+(a^6} \\
- \phantom{+(a^6}(a + a^2 + a^3 + a^4 + a^5 + a^6 + ...) \\
= 1.$$
