The base method in Vedic math is, to all intents and purposes, what you're looking for. A few decent sources would be here and here.
Roughly speaking, this is based upon the following idea. Suppose you have to multiply numbers $f$ and $g$ such that, after further transformation if necessary, $f$ and $g$ are very close to a number whose square is easy to multiply with other numbers (thus, typically powers of $10$, and their multiples like $25,50$ etc. are preferred). Such a number will be called the "base" $B$.
Suppose $B+p = f$ and $B+q = g$. Now, we write \begin{align*}
fg = (B+p)(B+q) = B^2 + Bp+Bq + pq &= B(B+p+q)+pq\\
& \boxed{= B(f+q) + pq}\\
& \boxed{= B(g+p) + pq}
\end{align*}
This formula is particularly wieldy when $B$ is a power of $10$, because if we, say, write out the calculation, then we obtain a digit-based break up \begin{align*}
&f && p \\
&g && q \\
\hline \\
&(f+q)|pq& &\text{ OR } (g+p)|pq
\end{align*}
Notice that $f+q=g+p$ is obtained by adding numbers crosswise to each other, while $|$ indicates "join as digits" : for example, $102\times 103$ would give $105|06 = 10506$ with base $100$. Of course, this appropriately adjusts if $p$ or $q$ is negative e.g. for $101\times 99$ we'd get $100|-1 = 10000 + (-1) = 9999$.
Your method is basically a usage of the appropriate modification when $B$ is not a power of $10$. For examples of this modification, see here or here.
The basic idea is to modify the $(f+q)$ or $(g+p)$ part appropriately. For example, suppose you do $41\times 41$, and you want to use base $50$ but don't know how to fit it into the above framework. What you do, is that you wish $50$ was $100$, which is $2$ times larger : so you instead do is find $x,y$ using base $100$, but divide the first part $(f+q)$ or $(g+p)$ by that compensatory constant ($2$ in this case).
For example, suppose we want to do $41 \times 41$ with base $50$ which would become \begin{align*}
41 && -9 \\
41 && -9 \\
32 &\quad\quad\quad|\quad\quad& 81
\end{align*}
This is not correct, but the point is the base is $50$, not $100$, so divide $32$ by the compensatory factor $2$ and you get $1681$, which is correct.
Another example : say you're doing $53 \times 46$. You write this as
\begin{align*}
53 && 3 \\
46 && -4 \\
49 &\quad\quad\quad|\quad\quad& -12
\end{align*}
Now, you want to "divide" $49$ by the compensatory factor, which is $100/50 = 2$ (because you wish the base was $100$ but it's in fact $50 = 100/2$). Of course $49$ is odd, but the rule says you can go into decimals (because you have zeros to play with on the LHS) so you do $49/2 = 24.5$, which after adding the two right zeros becomes $2450$. The final answer is then $2450-12 = 2438$.
And why should this work? Let's say that you took the base as $B$ but you wished that it was a number $B'$. Once you write $(f+q)|pq$, you want to join $(f+q)$ and $pq$ digitally together because this is what you'd typically do if $B$ was a power of $10$. However, you can't do that because you actually want to find $B(f+q)+pq$, not the number that would actually be produced by digital concatenation, which is $B'(f+q)+pq$. However, $B(f+q) = B'(f+q) \div \frac{B'}{B}$, so you know that modifying the left hand side by dividing it by the compensatory factor gives you the right answer, after rounding.
Furthermore, $B'$ will typically be chosen as the power of $10$ just above or below $B$, so that it's convenient to figure out what $B'(f+q)$ is (just $f+q$ shifted by how many ever zeros).
Now let's do $43\times 44$ using the compensatory base method.
\begin{align*}
43 && -7 \\
44 && -6 \\
37 &\quad\quad\quad|\quad\quad& 42
\end{align*}
The $37$ will turn into $18.5$ after division by the compensatory factor $2$, which leads to $1850+42 = 1892$.
Let's see : you got $x=7$ and $y=6$ in your method because you're using $50-a$ instead of $a-50$ in the base method. That is, $x= -p$ and $y=-q$.
In the given calculation, the crosswise calculation of either $f+q$ or $g+p$ divided by the compensation factor gives the result. In your case, realize that $m = 25-\frac{x+y}{2}$ satisfies $2m =50 - (x+y) = 50 -(50-a+50-b) = a+b-50$, while $f+q = f + g-50$ by definition of $q$ in the base method. So, both methods derive the same left hand side, while the right hand side is also the same since $x=-p$ and $y=-q$ implies that $xy = pq$. Meanwhile, you also recombine them in the same way, and your "adjustment" rule is also equivalent to what the base method adjustment rule advocates.
So what you've discovered is just the modified base rule, really!
However, as I rightly mentioned in my comment, you can modify this in multiple ways, by taking various bases $B'$ which may be above or below base $B$.
For example, suppose you want to do $34 \times 32$. You wish the base was $10$, not $30$. Just perform the subtractions with base $30$ for now, the compensatory factor being $\frac 13$ (so you have to divide by $\frac 13$ i.e. multiply by $3$).
\begin{align*}
34 && 4 \\
32 && 2 \\
36 &\quad\quad\quad|\quad\quad& 8
\end{align*}
Now $36$ is not correct. So you compensate by doing $36 \times 3 = 108$ and the final answer is $1088$.
Another example, say you're doing $2508 \times 2489$, and you want to do this with base $2500$, which you know is $10000/4$. So you work with base $2500$ as is, to get \begin{align*}
2508 && 8 \\
2489 && -11 \\
2497 &\quad\quad\quad|\quad\quad& -88
\end{align*}
The compensatory factor is $10000/2500 = 4$, so you do $2497/4$ which is a decimal, $624.25$ (Note : Mentally this is not difficult to see, since $2500/4 = 625$ is easy to do, and you just need to subtract $3/4$ from this). The final answer is then $$
6242500 - 88 = 6242412.
$$
So you can see that the method works for a variety of bases and coefficients, and your method is one of these multifarious methods.
Some more such methods in the literature include the criss-cross method of multiplication which works more generally.
