Original Question: Let $X$ and $Y$ be topological spaces and let $f:X \to Y$ and $g:X \to Y$ be isotopic embeddings. Is it true that $X \cup_f Y$ is homeomorphic to $X \cup_g Y$?
Edit: I meant to say the following:
Let $X$, $Y$ and $A \subset X$ be topological spaces and let $f:A \to Y$ and $g:A \to Y$ be isotopic embeddings. Is it true that $X \cup_f Y$ is homeomorphic to $X \cup_g Y$?
On
I believe the answer is no. Let $X=[0,1]$ be an interval with basepoint $0$ and let $Y$ be a wedge of two circles with basepoint $y$ at their wedge point. Let $f\colon\{0\}\to Y$ be given by $f(0)=y$ and let $g\colon\{0\}\to Y$ be given by $g(0)=y'$ for some $y'\neq y$.
I think cut point arguments show that $X\cup_f Y\ncong X\cup_g Y$ because the second space doesn't have a point which cuts the space into three path components but the first does.
No need to assume isotopy: Both spaces are homeomorphic to $Y$. Did you have a different question in mind?
Edit: Now that you have edited your question, here is a more interesting answer:
There exist $X, Y, A, f, g$ as in your question so that $X\cup_f Y$ is not even homotopy-equivalent to $X\cup_g Y$.
More specifically. Let $H$ denote the Hawaiian earring and let $a\in Y$ denote the point at which all the circles are attached (the origin) and $A=\{a\}$. Now, let $H', H''$ denote the subspaces of $H$ formed by all circles whose radii are of the form $1/(2n+1)$ and $1/2n$ respectively. It is easy to see that $H'$ and $H''$ are homeomorphic. Thus, we can think of $H$ as the wedge $H'\wedge_p H''$; in other words, $H=H'\cup_f H''$, where $f: A\to H''$ is the identity map.
Now, let's form another space, $J= H'\cup_g H''$, where $g: A\to H''$ is a map sending $a$ to a point $a'\ne a$. Clearly, the maps $f, g$ are isotopic. On the other hand, one of Eda's theorem shows that fundamental groups of $H$ and $J$ are not isomorphic.