Suppose I have two Hilbert spaces: $\mathcal{H}_1$ with the basis $\{| 0 \rangle_1, | 1 \rangle_1 \}$ and $\mathcal{H}_2$ with the basis $\{| 0 \rangle_2, | 1 \rangle_2 \}$, and I have a unitary $U$ acting on the composite Hilbert space $\mathcal{H}_1 \otimes \mathcal{H}_2$. Can I find a unitary $U^{'}$ acting on the Hilbert space $\mathcal{H}_1 \oplus \mathcal{H}_2$ spanned by the basis vectors $\{ |0 \rangle_1, | 1 \rangle_1, |0 \rangle_2, | 1 \rangle_2 \}$, such that the action on the composite Hilbert space is equivalent?
In other words, given an operator $U$ in $\mathcal{H}_1 \otimes \mathcal{H}_2$, find $U^{'}$ in $\mathcal{H}_1 \oplus \mathcal{H}_2$ such that
$$ \begin{bmatrix} |0 \rangle_1 \\ |1 \rangle_1 \\ |0 \rangle_2 \\ |1 \rangle_2 \end{bmatrix} \longrightarrow U^{'} \begin{bmatrix} |0 \rangle_1 \\ |1 \rangle_1 \\ |0 \rangle_2 \\ |1 \rangle_2 \end{bmatrix} \implies \begin{bmatrix} |0 \rangle_1 |0 \rangle_2 \\ |0 \rangle_1 |1 \rangle_2 \\ |1 \rangle_1 |0 \rangle_2 \\ |1 \rangle_1 |1 \rangle_2 \end{bmatrix} \longrightarrow U \begin{bmatrix} |0 \rangle_1 |0 \rangle_2 \\ |0 \rangle_1 |1 \rangle_2 \\ |1 \rangle_1 |0 \rangle_2 \\ |1 \rangle_1 |1 \rangle_2 \end{bmatrix} $$
After digesting Norbert Schuch's comments plus reading the OP carefully again plus reviewing some old notes I have to convert the previously accepted answer to a long comment:
We know that the "one-particle", aka single qubit, Hilbert space ${\cal H}$ describing only spin up/down is spanned by $\{|0\rangle,|1\rangle\}$ and may be identified with $\mathbb C^2$. The two-qubit Hilbert space can therefore be viewed as $\mathbb C^2\otimes\mathbb C^2$ which is spanned by all Kronecker products of two complex two-vectors, i.e., by all matrices $$ \begin{pmatrix}\varphi_1\psi_1&\varphi_1\psi_2\\\varphi_2\psi_1&\varphi_2\psi_2\end{pmatrix}\,,\quad\varphi_1,\varphi_2,\psi_1,\psi_2\in\mathbb C\,. $$
As a vector space $\mathbb C^2\otimes\mathbb C^2$ is obviously isomorphic to $\mathbb C^4=\mathbb C^2\color{red}{\oplus}\mathbb C^2\,.$ However, $\mathbb C^4=C^2\color{red}{\oplus}\mathbb C^2$ with its canonical scalar product $$ \langle \mu,\nu\rangle=\overline{\mu}_1\nu_1+\overline{\mu}_2\nu_2+\overline{\mu}_3\nu_3+\overline{\mu}_4\nu_4 $$ is not the right Hilbert space to describe the two-qubit system.
The right scalar product for $\mathbb C^2\otimes\mathbb C^2$ starts with the definition $$\tag{1} \langle \varphi\otimes\psi,\mu\otimes\nu\rangle:=\langle\varphi,\mu\rangle\langle\psi,\nu\rangle=(\overline{\varphi}_1\mu_1+\overline{\varphi}_2\mu_2)(\overline{\psi}_1\nu_1+\overline{\psi}_2\nu_2) $$ for product states which can be linearly extended to all states of $\mathbb C^2\otimes\mathbb C^2$ since they are linear combinations of product states. The fact that we norm all single qubit states to one is also important to make the scalar product well defined.
To see that $\mathbb C^2\otimes\mathbb C^2$ with (1) is the right Hilbert space, and not $\mathbb C^2\color{red}{\oplus}\mathbb C^2\,,$ consider three qubits. Their Hilbert space is -as we know- $\mathbb C^2\otimes\mathbb C^2\otimes\mathbb C^2$ which is eight-dimensional and not six-dimensional like $\mathbb C^6$.
To make a long story short, the coincidence that $\mathbb C^2\otimes\mathbb C^2$ is isomorphic to $\mathbb C^2\color{red}{\oplus}\mathbb C^2$ (as a vector space) led me to formulate the below answer which does not really address your issue.
Previous answer:
If you identify each Hilbert space ${\cal H}_i$ with $\mathbb C^2$ then ${\cal H}_1\otimes {\cal H}_2$ will become isomorphic to $\mathbb C^4$. A unitary $U'$ that acts only on ${\cal H}_1$ would be a matrix $$ U'=\begin{pmatrix}u_{11}&u_{12}& 0 & 0\\ u_{21}&u_{22}&0&0\\ 0&0&1&0\\0&0&0&1 \end{pmatrix}\,. $$ If you compose this with a unitary $U''$ that acts only on ${\cal H}_2$ you will end up with
$$ U''U'=\begin{pmatrix}u_{11}&u_{12}& 0 & 0\\ u_{21}&u_{22}&0&0\\ 0&0&u_{33}&u_{34}\\0&0&u_{43}&u_{44} \end{pmatrix}\,. $$ It is clear that not every unitary $U:\mathbb C^4\to\mathbb C^4$ can have this form.
In the language of tensor products: $U(2)\otimes U(2)$ is only a subgroup of $U(4)$ as mentioned in ZeroTheHero's comment.