Suppose your initial number has highest nontrivial factor $a$ and smallest nontrivial factor $b$. Then the devil gets to move by $\pm a$ and you get to move by $\pm b$ as well as multiplying and dividing by $b$.
You can certainly arrange
that the devil cannot win. To do this, make $a$ odd and $b$ even, by taking your initial number to be twice an odd number. Then the devil always has to make odd-sized moves and you can always make the number even (by multiplying by $b$, which will in fact be 2) before he does, so he can never move to the even numbers 0 or 100.
However,
this is not the same thing as actually winning because maybe the devil can keep the game going on for ever. Suppose he can't. Then you can somehow arrange to reach a position in which the devil's only available moves -- from $t$ to $t\pm a$, say -- are to positions from which you win immediately. (If you can't arrange that with best diabolical play, then the devil can simply always decline to move to a position from which you win immediately.) You can win by moving to $0$ from $\pm b$, or by moving to $100$ from $100\pm b$ or from $100b-\varepsilon$ where $0\leq\varepsilon<b$ or (if $b|100$) from $100/b$.
So, at the very least,
for you to be able to win even against a stupid devil there must be some $t$ for which both $t-a$ and $t+a$ are among these six numbers. We shall see that this is quite a tough condition to meet. Let's first get some boring cases out of the way. If you pick a prime number, so that $a=b=1$, then you can never change the number by more than 1 and so the devil can simply always move away from 0 or 100 if you get near them, and therefore avoid ever losing. (We have seen above that he can't hope to do better if you choose your initial number wisely.) If you pick the square of a prime, so that $a=b=p$, the same is true. (Note that $p^2\pm p$ can't be either 0 or 100.) So let us suppose that your starting number is not of either of these forms; in particular $1<b<a$. Note also that $b\leq9$ because it's the smallest nontrivial factor of a number $\leq90$, and that $a\leq45$ because it's a nontrivial factor of a number $\leq90$. Oh, and their product $ab$ is our original number, which is $\leq90$.
Now
let's suppose there's a $t$ for which both of $t\pm a$ are among those six numbers, and suppose for a moment that $100/b$ isn't one of them. We can put the others into groups I'll call A ($\pm b$), B ($100\pm b$), and C ($100b-\varepsilon$). The two numbers in group A, or the two in group B, can't be $t\pm b$ since $a\neq b$, so $t-a,t+a$ must be in different groups. They differ by $2a\leq90$. Any number in group C is at least 199, and unless $b=2$ is at least 298; this immediately means that group C is too far away from the other two groups. So we must have one number in group A and one in group B. These differ by at least $100-2b$ and therefore $2a\geq100-2b$ or $a+b\geq50$. But this isn't actually possible, because $a\leq45$ then implies $b\geq5$ which implies $a\leq18$ since $ab\leq90$.
It follows that
if such a $t$ exists, one of the numbers $t\pm a$ must be $100/b$ -- and, in particular, the latter must be an integer. So $b$ is one of $2,4,5$. We can't actually have $b=4$ because if $4|x$ then also $2|x$ so 4 wasn't the smallest factor after all. If $b=5$ then $6\leq a\leq18$ and in particular the gap from $100/b=20$ to $100\pm b$ is too large, so our two numbers must be $\pm5$ and 20 ... which differ by an odd number which therefore cannot be $2a$. So we can't have $b=5$ either and must have $b=2$.
So then
those six numbers are $-2, 2, 50, 98, 102, 200$ and we must have $2a\in\{48,52\}$, so either $a=24$ or $a=26$, corresponding to an initial choice of either 48 or 52. In either of these cases it is at least possible for the devil to get into a losing position; and in either case there are in fact exactly two such losing positions: $26,74$ if $a=24$ and $24,76$ if $a=26$.
And in these cases
in fact you can win. If you pick $x=48$ so that $a,b=24,2$ then the devil's first move must be to $24$ or $72$. Then you move to $26$ or $74$ respectively, whereupon the devil has to move to $2$ or $50$ (in the former case) or to $50$ or $98$ (in the latter) and you win. Similarly, if you pick $x=52$ so that $a,b=26,2$ then the devil must begin by moving to $26$ or $78$; then you move to $24$ or $76$ and again he has to move to $2$ or $50$, or to $50$ or $102$ and you win.
In conclusion:
You should begin by choosing either 48 or 52 as your number. In this case you can very quickly and easily force a win. If you choose any other number, then the devil cannot lose provided he never moves to a position from which you can win instantly (and there is no position from which he can't avoid that). But for at least some other numbers, you can keep him playing for ever without any risk of your losing.
