Since I know nothing about Wang tiles, my notation and terminology are probably bad. I hope my argument is correct.
Let $W$ be a solvable finite set of Wang tiles.
A set $\mathcal A$ of finite blocks is avoidable (with respect to the fixet set $W$ of Wang tiles) if there is a ($W$-)tiling of the plane in which no element of $\mathcal A$ occurs.
$\emptyset$ is avoidable, since $W$ is solvable.
A set $\mathcal A$ is avoidable if and only if every finite subset of $\mathcal A$ is avoidable.
Proof. For the nontrivial direction, suppose every finite subset of $\mathcal A$ is avoidable. Since $W$ is finite, $W^{\mathbb Z\times\mathbb Z}$ is a compact space. ($W$ has the discrete topology, $W^{\mathbb Z\times\mathbb Z}$ the Tychonoff product topology.) $F_0=\{T\in W^{\mathbb Z\times\mathbb Z}:T\text{ is a proper tiling }\}$ is a closed set, as is $F_A=\{T\in W^{\mathbb Z\times\mathbb Z}:A\text{ does not occur in }T\}$ for each $A\in\mathcal A$. Since the family $\{F_0\}\cup\{F_A:A\in\mathcal A\}$ has the finite
intersection property, it has nonempty intersection; i.e., $\mathcal A$ is avoidable.
Hence there is a maximal avoidable set of finite blocks, choose one and call it $\mathcal A$.
Let $S$ be a tiling of the plane in which no element of $\mathcal A$ occurs. I claim that every finite block occurring in $S$ occurs infinitely often. Assume for a contradiction that some finite block $B$ occurs at least once in $S$ but occurs only a finite number of times. Then $S$ contains arbitrarily large squares in which $B$ does not occur at all, i.e., arbitrarily large squares can be tiled with no occurrence of any element of $\mathcal A\cup B$. Now another easy application of compactness shows that the whole plane can be tiled with no occurrence of any element of $\mathcal A\cup\{B\}$; that is, the set $\mathcal A\cup\{B\}$ is avoidable. But $B\notin\mathcal A$ (since $B$ occurs in $S$), so this contradicts the maximality of $\mathcal A$.
I guess this is the same idea as Aaron Meyerowitz's answer but more verbose. Well, maybe somebody will find a verbose answer easier to follow.
P.S. If your background is engineering and physics, you may not like my use of topological compactness. You can give an equivalent proof using König's infinity lemma, that a finitely-branching infinite tree has an infinite branch.