Yes, this is true.
First approach
Let more generally $(\mathfrak{g}, [-,-]_1)$ be a Lie algebra over some field $K$.
If $\mathfrak{h}$ is a $K$-vector space of the same dimension as $\mathfrak{g}$ then there exists an isomorphism of vector spaces
$$
\varphi
\colon
\mathfrak{h}
\to
\mathfrak{g} \,.
$$
We can use this isomorphism of vector spaces to pull back the Lie bracket on $\mathfrak{g}$ to a Lie bracket on $\mathfrak{h}$.
More precisely, we set
$$
[x, y]_2
:=
\varphi^{-1}( [ \varphi(x), \varphi(y) ]_1 )
$$
for all $x, y \in \mathfrak{h}$.
This defines a Lie bracket on $\mathfrak{h}$, and the vector space isomorphism $\varphi$ becomes an isomorphism of Lie algebras
$$
\varphi
\colon
( \mathfrak{h}, [-,-]_2 )
\to
( \mathfrak{g}, [-,-]_1 ) \,.
$$
(This Lie bracket $[-,-]_2$ is the unique Lie bracket on $\mathfrak{h}$ which makes $\varphi$ into an isomorphism of Lie algebras.)
Suppose now that $K = \mathbb{R}$ and that the Lie algebra $\mathfrak{g}$ is of finite dimension $n$.
If $x_1, \dotsc, x_n$ is a basis of $\mathfrak{g}$ then we get an isomorphism of vector spaces
$$
\varphi
\colon
\mathbb{R}^n
\to
\mathfrak{g} \,,
\quad
e_i
\mapsto
x_i
\qquad
\text{for all $i = 1, \dotsc, n$,}
$$
where $e_1, \dotsc, e_n$ denotes the standard basis of $\mathfrak{g}$.
As explained above we can pull back the Lie bracket $[-,-]_1$ on $\mathfrak{g}$ to a Lie bracket $[-,-]_2$ on $\mathbb{R}^n$ such that $\varphi$ becomes an isomorphism of Lie algebras.
Second approach
There is also a less abstract but equivalent way of explaining the above procedure.
Suppose again that $(\mathfrak{g}, [-,-]_1)$ is a Lie algebra over a field $K$ and let $x_1, \dotsc, x_n$ be a basis of $\mathfrak{g}$.
We can then consider the structure constants of $[-,-]_1$ with respect to the basis $x_1, \dotsc, x_n$.
These are the unique coefficients $c_{ij}^k \in K$ with
$$
[x_i, x_j] = \sum_{k=1}^n c_{ij}^k x_k
$$
for all $i, j = 1, \dotsc, n$.
That $[-,-]_1$ is a Lie bracket on $\mathfrak{g}$ is equivalent to the conditions
\begin{equation}
c_{ii}^\ell = 0 \,,
\quad
c_{ij}^\ell + c_{ji}^\ell = 0 \,,
\quad
\sum_{k=1}^n
( c_{ij}^k c_{k \ell}^m + c_{j \ell}^k c_{ki}^m + c_{\ell i}^k c_{kj}^m ) = 0
\tag{1}
\end{equation}
for all $i, j, \ell, m = 1, \dotsc, n$.
If we are now given another $K$-vector space $\mathfrak{h}$ with basis $y_1, \dotsc, y_n$ then we can define a unique bilinear map
$$
[-,-]_2
\colon
\mathfrak{h} \times \mathfrak{h}
\to
\mathfrak{h}
$$
such that
$$
[y_i, y_j]
:=
\sum_{k=1}^n c_{ij}^k y_k
$$
for all $i,j = 1, \dotsc, n$.
The structure constants of $[-,-]_2$ with respect to the basis $y_1, \dotsc, y_n$ of $\mathfrak{h}$ are (by construction) the same as the structure constants of $[-,-]_1$ with respect to the basis $x_1, \dotsc, x_n$ of $\mathfrak{g}$.
It follows that $[-,-]_2$ is a Lie bracket on $\mathfrak{h}$ since the conditions $(1)$ are satisfied.
Moreover, there exists a unique linear map $\varphi$ from $\mathfrak{h}$ to $\mathfrak{g}$ with
$$
\varphi(y_i) = x_i
$$
for all $i = 1, \dotsc, n$, and this linear map is an isomorphism of Lie algebras from $(\mathfrak{h}, [-,-]_2)$ to $(\mathfrak{g}, [-,-]_1)$.
In the given situation we have $K = \mathbb{R}$ and can choose $\mathfrak{h} = \mathbb{R}^n$ and for $y_1, \dotsc, y_n$ the standard basis of $\mathbb{R}^n$.
Best Answer
There is a way to define the Lie bracket on the tensor product as follows. Suppose that $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:\mathfrak{g}_1\times \mathfrak{g}_2\longrightarrow \mathfrak{g}_1$ and $B_2:\mathfrak{g}_1\times \mathfrak{g}_2\longrightarrow \mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by $$ [g_1\otimes g_2, g_1'\otimes g_2']:= B_1(g_1,g_2)\otimes B_2(g_1',g_2')\quad \text{for } g_1,g_1' \in \mathfrak{g}_1 \text{ and } g_2,g_2' \in\mathfrak{g}_2.$$ For example if $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.