Reducing one variant of Hamiltonian path to another Define

$A = \{&lt;G,s,t&gt; :G$  is un directed graph that has a Hamilton path from $s$ to $t\}$  

$B = \{&lt;G&gt; :G$  is un directed graph that has a Hamilton path$\}$

I would like to show that $A \le_p B$. 

My attempt: 

Given $&lt;G,s,t&gt;$ , the reduction outputs $&lt;G'&gt;$ where $G'$ defines as follows: 

we take $v_s$ and $v_t$ s.t there is an edge $e_s$ between $s$ and$v_s$ and an edge $e_t$ between $v_t$ and $t$.

Now we define $G' = (V',E')$ where $V' = V \cup \{v_0 , v_1\}$ and $E' = (E-\{e' :e' \ne e_s , s\in e'\} -\{e' :e' \ne e_t , t\in e'\}) \cup \{ \{v_0,v \} :\{v,s\} \in E \} \cup  \{ \{v_1,v \} :\{v,t\} \in E \} \cup \{ \{v_0,s \}\ , \{v_1,t\} \}$.

Now, if $&lt;G,s,t&gt; \in A$  , let $s , u_1,\dots ,u_n ,t$ be the Hamilton path , then $s , v_0 , u_1 , \dots , u_n, v_1 ,t$ is a Hamilton path in $G'$. 

The other direction does not work for me , if $G'$ has a Hamilton path then I want to say that the "end points" must be around $s,t$ but i'm not sure how to do that.
The simplest reduction, which is similar to what you were trying but simpler, is to add two new vertices to the graph in the instance of A: a vertex $v_s$ connected to $s$, and a vertex $v_t$ connected to $t$. Any Hamiltonian path in the new graph will connect $v_s$ to $v_t$, and so will restrict to a Hamiltonian path connecting $s$ to $t$. Vice versa, any Hamiltonian path in the original graph connecting $s$ to $t$ extends to a Hamiltonian path in the new graph connecting $v_s$ to $v_t$.