Remarks on your calculations $\newcommand{Tr}{\operatorname{Tr}}$
\begin{align}
S(\rho_1\otimes\rho_2) = - \mathrm{Tr}\left[\rho_1\otimes\rho_2\; ln\left(\rho_1\otimes\rho_2\right)\right] &= -\sum_{\alpha,a}\langle 1:\alpha,2:a|\rho_1\otimes\rho_2|1:\alpha,2:a\rangle\log(\rho_1\otimes\rho_2)\\
\end{align}
I use now the fact that the logarithm of a product is equation to the sum of the logarithms: $\log(a b) = \log(a)+\log(b)$. Am I allowed to generalize this to the tensor product of two states?
The last equality does not make sense, as $\log(\rho_1\otimes \rho_2)$ is an operator : it should come before the ket (see my calculations below).
Given a hermitian operator $\hat \Lambda$ and a function $f$, we can define $f(\Lambda)$ the following way : if $|i\rangle$ is an orthonormal basis such that $\hat\Lambda|i\rangle = \lambda_i|i\rangle$, then we define $f(\lambda)$ as the linear operator such that :
$$f(\hat\Lambda) |i\rangle = f(\lambda_i) |i\rangle$$
(It turns out that this operator does not depend on the eigenbasis we chose.)
Since $f(\lambda) = \lambda \log(\lambda)$ is well-defined on $[0,+\infty)$ (by setting $f(0) = 0$, we can define $\hat \Lambda \log \hat \Lambda$ if $\hat \Lambda$ is positive semi-definite (which is true for density matrices).
Using bases of eigenvectors
Let $|i\rangle_1$ and $|x\rangle_2$ be orthonormal bases of eigen-vectors of $\rho_1$ and $\rho_2$ respectively, with :
$$\rho_1 |i\rangle_1 = p_i|i\rangle_1 \qquad \text{and}\qquad \rho_2|x\rangle_2 = q_x |x \rangle_2$$
Then, $|i\rangle_1 \otimes |x\rangle_2$ is a basis of $\mathcal H_1\otimes \mathcal H_2$, so we can use it to compute the trace :
\begin{align}
S(\rho_1\otimes \rho_2) &= -\Tr(\rho \log\rho) \\
&= -\sum_{i,x}\langle i|_1 \otimes \langle x|_2(\rho_1\otimes \rho_2\log(\rho_1\otimes \rho_2)) |i\rangle_1\otimes |x\rangle_2 \\
&= -\sum_{i,x} p_i q_x \log (p_i q_x) \\
\end{align}
From this point on, we are only dealing with real numbers, so we get :
\begin{align}
S(\rho_1\otimes \rho_2) &= -\sum_{i} p_i \log (p_i)-\sum_{x} q_x \log ( q_x) \\
&= S(\rho_1) + S(\rho_2)
\end{align}
What is $\log( \hat A\otimes \hat B)$ ?
In the intermediate steps of the calculation above, we showed that if $\hat A$ and $\hat B$ are positive definite hermitian operators on $\mathcal H_1$ and $\mathcal H_2$, then :
$$\log (\hat A \otimes \hat B) = \log(\hat A)\otimes \mathbb I_2 + \mathbb I_1 \otimes\log(\hat B) $$
where $\mathbb I_1$ and $\mathbb I_2$ are the identity operators on $\mathcal H_1$ and $\mathcal H_2$.
Using this formula, we can redo the calculations directly, without using a basis :
\begin{align}
S(\rho_1\otimes \rho_2) &= -\Tr(\rho_1\otimes \rho_2 \log(\rho_1 \otimes \rho_2) )\\
&= -\Tr(\rho_1\otimes \rho_2 \cdot( \log(\rho_1) \otimes \mathbb I_2 + \mathbb I_1 \otimes \log \rho_2) )\\
&= -\Tr ( (\rho_1 \log \rho_1) \otimes \rho_2 + \rho_1 \otimes (\rho_2 \log \rho_2)) \\
&= -\Tr ( \rho_1 \log \rho_1))\Tr( \rho_2) - \Tr(\rho_1)\Tr(\rho_2 \log \rho_2)) \\
&= S(\rho_1) + S(\rho_2)
\end{align}
Best Answer
The short version of your question is
It should be immediate that this implies what you ask.
To prove it, you just need the definition of the partial trace: $$ \mathrm{tr}_B(Z_{AB}) = \sum_B \langle i|_B Z_{AB}|i\rangle_B\ , $$ and the fact that $\mathrm{tr}_{AB}(Z)=\mathrm{tr}_A(\mathrm{tr}_B(Z))$.