Conditioning the ant to never pass by vertex $5$ is equivalent to cancelling edges $\{4,5\}$, $\{6,5\}$ and $\{8,6\}$. This leaves $7$ vertices and $9$ edges. The symmetry with respect to the plane containing $1$, $2$, $5$ and $8$ shows that $3$ and $7$ play the same role for a path starting from $1$ and ending at $8$. Likewise, $4$ and $6$ play the same role. So one can consider $3$ and $7$ as a single vertex, and $4$ and $6$ as a single vertex, only with more edges leading to them and leaving them than the others.
Thus the setting is equivalent to a random walk on a weighted graph with $5$ vertices and $5$ edges, drawn like a square plus an additional edge added to a vertex of the square. The square has vertices $1$, $2$, $3$ (which represents $3$ and $7$) and $4$ (which represents $4$ and $6$), the fifth vertex is $8$, and the edges are those of the square $\{1,2\}$, $\{2,3\}$, $\{3,4\}$, $\{4,1\}$, plus the additional edge $\{3,8\}$. Every edge has weight $2$ except the edge $\{1,2\}$ which has weight $1$. Weights on edges mean for instance that starting from $1$, one has $1/(1+2)$ chances to go to $2$ and $2/(1+2)$ chances to go to $4$.
Writing $t_i$ for the mean hitting time of $8$ starting from $i$ on the reduced graph, one looks for $t_1$. The vector $(t_1,t_2,t_3,t_4)$ solves the system $t_1=1+(t_2+2t_4)/3$,
$t_2=1+(t_1+2t_3)/3$, $t_3=1+(t_1+t_2+0)/3$ and $t_4=1+(t_1+t_3)/2$. Hence $t_1=62/5$ or something like that.
Alternatively, one can solve an analogous system on the original graph with $7$ vertices and $9$ edges. This system has size $6$ since $t_8=0$ and if one solves it one will note that some coordinates of the solution are equal, namely $t_3=t_7$ and $t_4=t_6$. The reduction explained above uses this fact to lower a priori the size of the linear system down to $4$ but it is not essential.
All this, and much more, in the must-read short book Random walks and electric networks by Peter G. Doyle and J. Laurie Snell.
Your chain is wrong because the $6$ other vertices are different in that $3$ of them are adjacent to $x$ and $3$ are adjacent to $y$.
The answer is actually independent of the particular graph and holds for any pair of vertices in any irreducible Markov chain.
Let $p$ be the probability that the chain reaches $y$ before returning to $x$. The walk goes to $y$ with probability $p$, and then it keeps trying to return to $x$ with success probability $p$, which is expected to take $\frac1p$ tries. Thus the expected number of visits to $y$ before returning to $x$ is $p\cdot\frac1p=1$.
You can also see this from looking at the sequence of states in the long term. Since the two corners are visitied equally often, between any two instances of one there must on average be $1$ instance of the other.
The second argument works for any pair of states not necessarily related by symmetry, whereas in the first argument it’s not immediately obvious that the probability $p$ is the same in the two directions unless, as in this case, the two vertices are related by symmetry.
Best Answer
Consider the possible locations of the ant after the first two moves. The ant can be on vertex A in 3 path(A->D->A,A->C->A,A->B->A), on vertex B in 1 path(A->C->B), on vertex C in 1 path(A->B->C), vertex E in 2 paths(A->D->E,A->B->E), vertex F in 2 paths(A->D->F,A->C->F) and D in 0 paths.
Now we do the reverse: from the above vertices back to A, in two moves. For each vertex, the possible paths would just be the paths of the first two moves in reverse order(for Ex: from vertex E back to A would correspond to E->D->A, E->B->A).
The require probability is $\frac{3^2+1^2+1^2+2^2+2^2}{3^4}=\frac{19}{81}$