Update: Thanks to @Ken for pointing out my mistake in the comments. Here is a corrected answer.
If you want the distribution to be up to homeomorphism, then the smash product distributes over finite wedges for arbitrary based spaces. This doesn't work the same for infinite wedges in the regular topological category. However, in the general case, there is a canonical continuous bijection: $f:\bigvee_{j\in J}X\wedge Y_j\to X\wedge \bigvee_{j\in J}Y_j$. This bijection is a weak homotopy equivalence but may not always be a true homeomorphism. Algebraic topologists often deal with this difficulty by working internal to a convenient category such as the category of $k$-spaces.
Proof of distribution over finite wedges. Let $(X,x_0)$ and $(Y_j,y_j)$, $j\in J$ be based spaces where $J$ is finite. Let $w_0$ be the wedge point of $\bigvee_{j\in J}Y_j$. Consider the following commutative diagram of canonical quotient maps.
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{ccc}
X\times \coprod_{j\in J}Y_j & \ra{\cong} & \coprod_{j\in J}X\times Y_j \\
\da{} && \da{} \\
X\times \bigvee_{j\in J}Y_j & \ra{p} & \bigvee_{j\in J}X\times Y_j\\
\da{} && \da{\vee_{j}q_j} \\
X\wedge \bigvee_{j\in J}Y_j & & \bigvee_{j\in J}X\wedge Y_j
\end{array}
$$
The top map is clearly a homeomorphism and we can identify these two spaces if we like. The middle horizontal map $p$ collapses $X\times \{w_0\}$ to the wedge point of $\bigvee_{j\in J}X\times Y_j$. If $q_j:X\times Y_j\to X\wedge Y_j$ is the canonical quotient, then the vertical map $\vee_{j}q_j:\bigvee_{j\in J}X\times Y_j\to \bigvee_{j\in J}X\wedge Y_j$ is quotient because wedges of based quoitient maps are quotient. Since all the maps are quotient maps, the universal property of quotient spaces tells you that have have the desired homeomorphism $X\wedge \bigvee_{j\in J}Y_j\cong \bigvee_{j\in J}X\wedge Y_j$ that makes the diagram commute if it holds that the two vertical compositions make the same identifications. If you check both of them, all fibers are points except that the fiber over each basepoint is $\left(X\times\coprod_{j\in J}\{y_j\} \right)\sqcup \left(\{x_0\}\times\coprod_{j\in J} Y_j\right)=\coprod_{j\in J}(X\times \{y_j\}\cup \{x_0\}\times Y_j)$. $\square$
Note: the intermediate quotient maps are not necessary to the proof but might help you keep track of things by breaking up the construction into steps.
In the general case (when $J$ is infinite), it's best to assume the spaces involved are Hausdorff. The right composition in the above diagram is quotient, but the left composition might not be quotient because the wedge $\bigvee_{j}Y_j$ will not always be locally compact Hausdorff. Hence, we get a continuous bijection $f:\bigvee_{j\in J}X\wedge Y_j\to X\wedge \bigvee_{j\in J}Y_j$ but this may not be a homeomorphism. However $f$ maps any compact subspace $K$ of $\bigvee_{j\in J}X\wedge Y_j$ homeomorphically onto $f(K)$ (we're using the closed mapping theorem here, which is why we want the spaces to be Hausdorff). This implies that $f$ is actually a bijective weak homotopy equivalence. In fancier categorical terms, it means that the two spaces have homeomorphic $k$-space coreflections, $k(\bigvee_{j\in J}X\wedge Y_j)\cong k(X\wedge \bigvee_{j\in J}Y_j)$.
