$\gcd(x,y)=1$ $\implies$ there are integers $s,t$ such that $sx+ty=1$.
Let $z=ax=by$. Then $sx+ty=1$ $\implies$ $sxz+tyz=z$, i.e. $(sb+ta)xy=z$. Thus $xy\mid z$.
As in the pair case, let's first find the number of ordered triples without restrictions on the order of the entries.
Every prime factor $p_i$ with multiplicity $a_i$ occurs in at most two divisors.
There is $1$ way for $p_i$ to occur in zero divisors.
There are $3a_i$ ways for $p_i$ to occur in one divisor.
There are $3a_i^2$ ways for $p_i$ to occur in two divisors.
Thus, in total there are $3a_i^2+3a_i+1$ ways to distribute $p_i^{a_i}$ over the divisors, and hence there are $\prod_i(3a_i^2+3a_i+1)$ ordered triples. From this we want to deduce the number of unordered triples (or, equivalently, the number of ordered triples with the order restrictions on the entries).
There is one triple with three equal entries, namely $(1,1,1)$. There are $\prod_i(2a_i+1)-1$ ordered pairs of distinct coprime divisors, each of which corresponds to three ordered triples with two equal entries but only one unordered triple with two equal entries. Thus the count of unordered triples is
$$
\frac{\prod_i(3a_i^2+3a_i+1)-3\left(\prod_i(2a_i+1)-1\right)-1}6+\prod_i(2a_i+1)-1+1=\frac{\prod_i(3a_i^2+3a_i+1)+3\prod_i(2a_i+1)+2}6\;.
$$
You can check that your examples all come out right.
Best Answer
The sum is relatively prime under all conditions.
Actually for your problem the only two prime factors of $10^x$ are $2,5$. And the only two factors of $p$ is $p$ and $1$. So do either $p,2,5$ divide $p + 10^x$?
But for a MUCH more general and more useful result, read on:
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If $W$ and $V$ have no factors (other than $1$) in common. And if $d\ne 1$ is a factor of $W$ then $d|W$ and $d\not \mid V$. So $W+V = d(\frac Wd + \frac Vd)$. $\frac Wd$ is an integer. $\frac Vd$ is not. So $d$ is not a factor of $W+V$.
So $W$ and $W +V$ have no non-zero factors in common.
....
Actually we can take it a step further that $\gcd(W,V) = \gcd(W, W \pm k*V) = \gcd(V, V \pm j*W)$ for any integers $j,k$. (But don't fall into the trap that $\gcd(V, W)= \gcd (kW \pm jV, lW\pm mV)$. That is simply false. Utterly false.)
I'll leave the proof of that as a research project.