In your equation, if you solve for $a$ you get:
$$a = \frac{\sqrt{M^2-60b^2}}{3} - 4b$$
If we're looking for integer solutions then this means that $M^2-60b^2$ must be a square. Also it must be divisible by $3$, so at least you only need search over (pseudo-)triples of the form: $$9x^2 + 60b^2 = M^2$$
Moreover you want $a$ to be positive so you require $x \ge 4b$. There is only one primitive triple with this property which yields $\{a,b,M\} = \{2,3,48\}$. Any multiple of this triple would also work ($\{4,6,96\}, \{6,9,144\}, \ldots$).
You can reduce the effort to a level that can be handled without a computer.
Let's write the calculation as
\begin{array}{cc}
&A&B\\
\times&&C\\\hline
&D&E\\
+&F&G\\\hline
&H&I
\end{array}
There are strong restrictions on $A$ and $C$, since their product must be single-digit. We can't have $C=1$, since otherwise $DE=AB$. So either $A=1$, or $(A,C)$ must be one of $(2,3)$, $(2,4)$, $(3,2)$ or $(4,2)$. We can exclude $(2,4)$ and $(4,2)$: This implies $D=8$, $F=1$ and $H=9$, but then adding $E$ and $G$ mustn't result in a carry, which is impossible, since $I$ can't be $8$ or $9$ and $E+G$ must be at least $3+5$.
So unless $A=1$, $(A,C)$ must be either $(2,3)$ or $(3,2)$. Also $B\notin\{1,5\}$. That leaves $10$ possible triples for $(A,B,C)$, which each determine values for $D$ and $E$, and it shouldn't be too much effort to go through those $10$ cases, find the $4$ remaining options for $F$, $G$, $H$ and $I$ and show that they don't work out.
That leaves us with $A=1$. In that case the options for $(C,B)$ can be enumerated as $(2,7)$, $(2,8)$, $(2,9)$, $(3,4)$, $(3,6)$, $(3,8)$, $(3,9)$, $(4,3)$, $(4,7)$, $(4,8)$, $(4,9)$, $(6,3)$, $(6,7)$, $(6,9)$, $(7,2)$, $(7,4)$, $(7,6)$, $(7,8)$, $(7,9)$, $(8,2)$, $(8,3)$, $(8,4)$, $(8,7)$, $(8,9)$. That's $24$ more cases, for a total of $34$ cases to be checked.
Best Answer
The strategy I would use is the following:
Goal is to get $100$ as sum. Among the digits $123456789$, pick and choose the sum close to $100$, such as $89$. Therefore I would attempt to get a value of $11$ from $1234567$ using different combinations.
When you start working on a smaller sum now (sort of like divide and conquer), you may get the desired result. (Of course there is no specific algorithm).
In order to get $11$, I have
$$(1\times 23)-4+5-6-7 = 11$$
$$(1-2+3-4+5)\times 6 -7= 11$$
$$123-45-67 = 11$$
Therefore
$$(1\times 23)-4+5-6-7+89 = 100$$
$$(1-2+3-4+5)\times 6 - 7+89=100$$
$$123-45-67+89=100$$
${\bf{Adding}}$ ${\bf{more}}$ to it: If we look at $78+9 = 87$ and instead of $89$, we seek the remaining $13$ to be derived from $123456$, and one way to get that is
$$6+5+4-3+2-1=13$$
Therefore
$$78+9+6+5+4-3+2-1=100$$