Label a state (where it is your turn) by $(n,m)$ where $n$ is the number of stones in the pile and $m$ is the parity ($0= even$, $1=odd$) of the number of stones you have. Note that the total number of stones is odd, so if
$n$ is even your opponent's number of stones has the opposite parity to yours, while if $n$ is odd your opponent's number has the same parity as yours.
Thus $(0,0)$, $(1,1)$ are winning positions and $(0,1)$ and $(1,0)$ are losing.
By induction, I find the following:
$(4k,0)$ is a win, $(4k,1)$ is a loss
$(4k+1,0)$ is a loss, $(4k+1,1)$ is a win
$(4k+2,0)$ is a win, $(4k+2,1)$ is a win.
$(4k+3,0)$ is a win, $(4k+3,1)$ is a loss.
Thus from $(4k,0)$ (if $k>0$) you take one stone, leaving your opponent in the losing position $(4k-1,1)$.
From $(4k,1)$ you have a loss, as you must leave your opponent with either $(4k-1,0)$ or $(4k-2,1)$, both wins.
Similarly in other cases.
Almost the same strategy works: suppose that player $1$'s first move is to pick one stone. Then there are exactly $100$ stones left, and at this point player one cane make use the strategy that you were describing above. Since $100$ is divisible by $5$, player $1$ is sure to win the game.
Best Answer
This is equivalent to a variant of Nim. In this variant you have piles $P_1,\ldots,P_n$ for some $n$. A move consists in choosing a pile $P_k$ with $k>1$ and transferring any positive number of stones from $P_k$ to $P_{k-1}$, or removing any positive number of stones from $P_1$. The first person who has no valid move loses. Equivalently, the person who takes the last stone off the board wins.
I claim that this game is equivalent to ordinary Nim with piles $P_{2k-1}$ for $k=1,\ldots,\lceil n/2\rceil$, i.e., with the odd-numbered piles of the original variant game. Suppose that I leave my opponent with a position in which the nim sum of the sizes of the odd-numbered piles is zero. If he moves stones from pile $P_k$ to pile $P_{k-1}$, where $k$ is even, I move the same number from $P_{k-1}$ to $P_{k-2}$ (or off the board, if $k=2$), again presenting him with a position in which the nim sum of the sizes of the odd-numbered piles is zero. If he moves stones from pile $P_k$ to pile $P_{k-1}$, where $k$ is odd, the nim sum of the odd-numbered piles is no longer $0$. Thus, there is an odd-numbered pile $P_k$ from which I can remove stones to make the nim sum of the odd-numbered piles zero again, and I simply move those stones to $P_{k-1}$ (or off the board, if $k=1$). In short, no matter how he moves from such a position, I can present him with such a position on his next move.
In the traditional terminology, positions in which the nim sum of the sizes of the odd-numbered piles is zero are P-positions, and all other positions are N-positions. The winning strategy is to ignore the even-numbered piles and to make what would be the winning move in a game of ordinary Nim with just the odd-numbered piles.