Yes, you can conclude that.
First, by the defining relation of conditional probability $P(X \in A | Y)$, for any Borel set $B$, we have
\begin{align*}\int_{\{Y \in B\}}P(X \in A | Y) dP = P(X \in A, Y \in B). \tag{1}\label{1}\end{align*}Next, the given condition $P(X \in A | Y) = P(X \in A)$ with probability $1$ entails that the left hand side of $\eqref{1}$ reduces to\begin{align*}\int_{\{Y \in B\}}P(X \in A | Y) dP = \int_{\{Y \in B\}}P(X \in A)dP = P(X \in A)P(Y \in B). \tag{2}\label{2}\end{align*}$\eqref{1}$ and $\eqref{2}$ together then give $P(X \in A, Y \in B) = P(X \in A)P(Y \in B)$ for any Borel sets $A$ and $B$, which is the definition of independence of $X$ and $Y$.