1. Introduction
2. Easy example
3. Normality, Duality
4. Normal example
Introduction
The pseudomanifold and homology manifold conditions are both local conditions, while Poincaré duality is a global condition. It is possible for a pseudomanifold to fail the homology manifold conditions in several places, but so that the local deviations globally cancel, so it retains duality.
Simplicial complexes are "locally cone-like": every point $x$ has a neighborhood that is the cone on another space $L$, the link, with $x$ the cone point. Then $$H_*(X,X-\{x\})=H_*(CL,L\times I)=\tilde H_*(SL)=\tilde H_{*-1}(L)$$ So the homology manifold condition is that the links have the homology of spheres.
Easy example
The interval admits an involution reversing the endpoints. The quotient by this involution is again contractible. Think of this quotient operation as gluing one half of the interval to the other half. Embed the interval in an $n$-manifold. Form a quotient of the manifold by gluing the one half of the interval to the other half. If $n\geqslant 3$, this leaves the cells in degree $n$ and $n-1$ unchanged, so does not affect the pseudomanifold condition. Since we have replaced one contractible subspace by another, the homotopy type of the total space is unchanged and satisfies duality (with the pseudomanifold fundamental class, etc). But the the links are no longer spheres at any point along the interval. At the image of the ends of the interval, the link is two spheres glued in one point. At general points along the interval, it is two spheres glued in two points. At the image of the midpoint, the fixed point of the involution, the link is a single sphere with two of its points glued to each other.
Normality, Duality
That example involved changing a manifold in low dimensions, which doesn't violate the high dimensional pseudomanifold condition. To rule out such examples, one has the concept of a normal pseudomanifold (cf normal variety). The normalization of a pseudomanifold is produced by taking a disjoint union of $n$-simplices parameterized by those of the original space and gluing them along their $n-1$-faces, according to how the original was glued. Thus the normalization of the above example is the unmodified manifold. A normal pseudomanifold is one isomorphic to its normalization.
It may be useful to use the Verdier dualizing sheaf. The cohomology with coefficients in the dualizing sheaf matches the homology: $H^*(X; D)\cong H_*(X)$. An oriented pseudomanifold yields a map of sheaves $\mathbb Z\to D$. We seek Poincaré duality, that is, $H^*(X; \mathbb Z)\cong H_*(X)$. So we seek spaces where the map of sheaves induces an isomorphism on cohomology $H^*(X; \mathbb Z)\cong H^*(X; D)$, even though it is not an isomorphism of sheaves. That is, we seek the cone to be a nontrivial sheaf with no cohomology in any dimension. There are such sheaves, such as a local system supported on a circle with appropriate monodromy. That motivates the construction. It also gives immediate proofs of the claims, but it should not be hard to check them without using sheaves.
Normal example
Take an $n$-manifold manifold $M$ and an automorphism $\phi$ so that the action of on the homology in intermediate degrees is sufficiently mixing so that if we consider it an action of the group $\mathbb Z$ and take group homology with those coefficients, the homology is trivial. For example, $M=T^2$ and $\phi=\left(\begin{matrix}2&1\\1&1\\ \end{matrix}\right)$. Then form the mapping torus ($T_\phi=M\times I/(x,0)\sim(\phi(x),1)$). This is an $M$-bundle over the circle, homology computed by a spectral sequence $H_*(S^1;H_*(M))\Rightarrow H_*(T_\phi)$. By assumption on $\phi$, the spectral sequence degenerates and $H_*(T_\phi)=H_*(S^1\times S^n)$. Then cone off each copy of $M$, forming a circle of cone points. To put it another way: form the mapping cylinder of the map to the circle $T_\phi\to S^1$. Yet another way: the mapping torus of the self-map of the cone $\tilde\phi\colon CM\to CM$. This space is a normal pseudomanifold (with boundary) because those properties are preserved by cones and products. Its set of singular points is a circle, and the dualizing sheaf twists about it with the prescribed monodromy, so it contributes nothing to the sheaf cohomology. If you prefer a closed example, double the space along the boundary (or equivalently take the double mapping cylinder of the map to the circle; or the mapping torus of the automorphism of the suspension). This gives a pseudomanifold that is not a homology manifold, but which satisfies Poincaré duality. Just as the mapping torus had the homology of a manifold $S^1\times S^n$, this space has the homology of $S^1\times S^{n+1}$. Actually, that is only correct if we restrict to trivial coefficients. If we define Poincaré duality to be for all local systems, then this space fails, for some unwind the twist around the singular circles. However, we can eliminate them by killing the fundamental group by surgery. That is, cut out a neighborhood of a circle, $S^1\times D^{n+1}$, leaving a boundary $S^1\times S^n$ and fill it in with $D^2\times S^n$. Now local systems are trivial, so it satisfies full Poincaré duality.
Best Answer
Let $C = S^3 \setminus A$. Alexander duality says that $$ H_1(C) \cong H^1(A) \cong \Bbb Z\, . $$ Let $\alpha: S^1 \to C$ be any map representing a generator of $H_1(C)$ (every first homology class is spherical by the Hurewicz theorem). By homotopical approximation we can assume $\alpha$ is a smooth embedding. We can also assume without loss in generality that the image of $\alpha$ misses the north pole of $S^3$. Identify the complement of the north pole with $\Bbb R^3$. Set $B= \alpha(S^1)$. Then $A\amalg B\subset \Bbb R^3$. The degree of the map $\require{AMScd}$ \begin{CD} \ell: A \times B @>>> S^2 \end{CD} given by $(x,y) \mapsto (x - y)/|x - \alpha(y)|$ is the linking number of $A$ with $B$ by definition. On the other hand, this map has degree one (after choosing appropriate homology generators).
The point is that the pushforward of a generator \begin{CD} H_1(S^1) @>\alpha_\ast >> H_1(C) \end{CD} coincides with the degree of $\ell$.
How can we check this? Well, assuming $A$ has a nice regular neighborhood, we could redefine $C$ as the complement of that neighborhood. Then $\ell$ can be redefined as the degree of the map $$ A \times C \to S^2 $$ again given by the same formula, where we are assuming our new $C$ misses the north pole of $S^3$. Alexander duality says that the induced slant product pairing $$ H_1(A) \otimes H_1(C) \to H_2(S^2) = \Bbb Z $$ is non-singular, so the degree is $\pm 1$.
Even if $A$ fails to have a nice regular neighborhood, we can assume it misses the north pole $x$ and that $C:= S^n \setminus A$ misses another point $y$ of $S^n$. Identify $S^n \setminus x \cong \Bbb R^n \cong S^n \setminus y$. Then, similarly, we obtain a map $$ A\times C \to S^2 $$ and there a linking map $A\times B\to S^2$. With these changes, the argument proceeds as before.