Proof of obtaining multiple of 10 by using arithmetic operations on any three numbers

Let's collect some of the useful observations mentioned and implicit in the OP:

1. Only the residue class of $a,b,c$ mod $10$ matters.
2. If any of $a,b,c$ is divisible by $5$ then there is a solution.
3. If any subset of $\{a,b,c\}$ has a solution, then so does $\{a,b,c\}$.

As a consequence of the above, we may assume WLOG that $\{a,b,c\}$ are distinct single digits from $1,2,3,4,6,7,8,9$.

With a little more effort, we can justify that $a$ can be replaced by $-a$, which will let us further restrict to the digits $1,2,3,4$. It's not too hard to be convinced of this by playing around with some examples, but let's give a proper proof by structural induction. More specifically, we will prove:

Proposition: Let $f(x_1,\ldots,x_n)$ be a fully parenthesized expression using each $x_i$ exactly once and only $+,-,*$ operators. Then at least one of $-f$ or $f$ can be written as a similar expression using $-x_1,x_2,\ldots,x_n$ exactly once.

This is trivial for $n=1$, and for $n=2$ we quickly verify each of the four possible expressions of two variables: $$-(a+b) = (-a)-b,\\-(a-b) = (-a)+b,\\(b-a) = b+(-a),\\-(a*b) = (-a)*b.$$

We can now do a structural induction: suppose $n\ge 3$. By extracting the highest-level operator of $f(x_1,\ldots, x_n)$, we may write $f$ as $h(g_1(\cdots),g_2(\cdots))$, where each of $g_1,g_2,h$ are valid expressions, and the arguments of $g_1, g_2$ partition the set $\{x_1,\ldots,x_n\}$. I've deliberately obscured the arguments because the notation becomes unwieldy, as the idea is better illustrated by a simple example:

If $f(x_1,x_2,x_3,x_4,x_5) = ((x_1 * x_5) + x_2) - (x_3 - x_4)$, then we take $$h(a,b) = a-b,\quad g_1 = (x_1 * x_5) + x_2,\quad g_2 = x_3 - x_4.$$

Note that $h,g_1,g_2$ each take $<n$ arguments so we could apply the inductive hypothesis to any of them as desired. Now, $x_1$ belongs to exactly one of $g_1$ or $g_2$. By adjusting $h$, we can assume WLOG that it belongs to $g_1$. By hypothesis, either $-g_1$ or $g_1$ can be written in terms of $-x_1$ and the remaining arguments of $g_1$. If it is $g_1$ that may be so written, then we are done since $f=h(g_1,g_2)$ can be written in terms of $-x_1,x_2,\ldots,x_n$. Else, apply the induction hypothesis again to $h$ to see that either $f=h(g_1,g_2)$ or $-f=-h(g_1,g_2)$ can be written in terms of $-g_1$ and $g_2$, and thus also in terms of $-x_1,x_2,\ldots,x_n$.

The above proposition lets us freely replace $9$ by $-9$ (which is equivalent to $1$), etc., and so we can narrow down the set $\{a,b,c\}$ to a subset of $\{1,2,3,4\}$. Lucky for us, there's only four such subsets:

$$ (1 + 2 - 3) = 0,\\ (1 + 4)*2 = 10,\\ (1 + 3 -4) = 0,\\ (2+3)*4 = 20.$$

If you have directly verified this for all triples of one-digit numbers, then you have proved it. Because you only care about an arithmetic combination that makes $0$ mod $10$, so proving it for one-digit numbers proves the greater claim.

Without a direct verification, consider all $10^3$ triples of one-digit numbers. If $0$ is in the triple, then multiplying all three is a multiple of $10$.

So now consider all $9^3$ triples from 1--9. If $5$ is in the triple, then either both of the other numbers are odd and you can multiply $5(\text{odd}+\text{odd})$ to get a multiple of $10$; or at least one of the others is even and you can multiply all three to get a multiple of $10$.

So now consider all $8^3$ triples from 1--4,6--9. If any number appear twice in the triple, the difference is $0$, and that difference times the third number is a multiple of $10$.

So now consider all $\binom{8}{3}$ triples from 1--4,6--9 without repetition. If a number and its complement mod $10$ are both in the triple, then sum those and multiply by the third to get a multiple of $10$.

So now consider all $\binom{4}{3}\cdot2^3$ triples from 1--4,6--9 without repetition where no two numbers sum to $10$. We are down to only $32$ triples to consider, and inspecting them directly is not so bad. First, consider the $16$ triples with two or three members from 1--4.

$$\underline{12}3\ \color{magenta}{\underline{1}2\underline{4}}\ \color{orange}{\underline{1}2\underline{6}}\ \color{blue}{127}\ \underline{13}4\ \color{blue}{136}\ 1\underline{38}\ 1\underline{47}\ \color{magenta}{\underline{14}8}\ \color{magenta}{\underline{23}4}\ \color{magenta}{\underline{23}6}\ 2\underline{39}\ \color{orange}{\underline{2}4\underline{7}}\ \color{orange}{2\underline{49}}\ \color{orange}{\underline{3}4\underline{8}}\ 3\underline{49}$$

Blue sum to $0$ mod $10$.

Magenta have an (underlined) pair that sum to $5$ (or $15$) and the third number is even, so that sum times the third number is a multiple of $10$.

Orange have an (underlined) pair whose difference is $5$ and the third number is even, so that difference times the third number is a multiple of $10$.

The remaining black all have an (underlined) pair that sums to the third (mod $10$) so adding that pair and then subtracting the third makes a multiple of $10$.

Note that if we negate all members of one of these triples, we get the triples with two or more in 6--9. The same operations give a multiple of $10$ since we only ever either add and subtract [with the blue and black], or add/subtract and then multiply by an even number [with magenta and orange]. So we've indirectly checked all $32$ of the lingering cases.

Note: This answer has the focus to reduce the number of variants which are to study by consequently using modular arithmetic.

• $a,b,c\in\mathbb{Z}$

We are looking for multiples of $10$ when doing addition, subtraction and multiplication. Since we have \begin{align*} (a\, \mathrm{mod}\, (10)) + (b\, \mathrm{mod}\, (10)) &\equiv (a+b)\, \mathrm{mod}\, (10) \\ (a\, \mathrm{mod}\, (10)) - (b\, \mathrm{mod}\, (10)) &\equiv (a-b)\, \mathrm{mod}\, (10) \\ (a\, \mathrm{mod}\, (10)) \cdot (b\, \mathrm{mod}\, (10)) &\equiv (a\cdot b)\, \mathrm{mod}\, (10) \\ \end{align*}

it is sufficient to consider $\color{blue}{a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}}$.

• $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}$

If two of the three numbers are congruent modulo $10$, let's say $a\equiv b\, \mathrm{mod}\, (10)$ we obtain \begin{align*} \color{blue}{(a-b)\cdot c}\equiv 0\cdot c\color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}

and we are done. In the following we may WLOG assume: $a<b<c$

• $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\},a<b<c$

If one of them, let's say $a$ is zero we obtain \begin{align*} \color{blue}{a\cdot b\cdot c}&\equiv 0\cdot b\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}

and we are done.

• $a,b,c\in\{1,2,3,4,5,6,7,8,9\}, a<b<c$

If one of them, let's say $a$ is equal to $5$ we consider two cases.

First case: One of $b$ or $c$ is even. Let's assume $b$ is even. It follows \begin{align*} a&\equiv 0\, \mathrm{mod}\, (5)\\ b&\equiv 0\, \mathrm{mod}\, (2)\\ ab&\equiv 0\, \mathrm{mod}\, (10)\\ \end{align*} and we obtain \begin{align*} \color{blue}{a\cdot b\cdot c}&\equiv 0\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}

Second case: Both, $b$ and $c$ are odd. It follows from

\begin{align*} b\equiv 1\, \mathrm{mod}\, (2)\\ c\equiv 1\, \mathrm{mod}\, (2)\\ (b+c)\equiv 0\, \mathrm{mod}\, (2)\\ \end{align*}

and we obtain \begin{align*} \color{blue}{a\cdot (b+c)} \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}

• $a,b,c\in\{1,2,3,4,6,7,8,9\}, a<b<c$

Since

\begin{align*} 1\equiv (1-10)\equiv (-9)\, \mathrm{mod}\, (10)\\ 2\equiv (2-10)\equiv (-8)\, \mathrm{mod}\, (10)\\ 3\equiv (3-10)\equiv (-7)\, \mathrm{mod}\, (10)\\ 4\equiv (4-10)\equiv (-6)\, \mathrm{mod}\, (10)\\ \end{align*}

we can restrict the attention to $\color{blue}{a,b,c\in\{1,2,3,4\},a<b<c}$.

• $a,b,c\in\{1,2,3,4\}, a<b<c$

This case is already nicely considered in the answer from @ErickWong.

Conclusion: Doing arithmetic with three integers $a,b,c\in\mathbb{Z}$, each of them occurring once and using one or more of the operations addition, subtraction, multiplication and negation ($a \rightarrow -a$) we can always obtain an integer value which is a multiple of $10$.