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I'm adding my comment as a partial answer, as discussed there; this is a reformulation of point 2 that has been added to the question.

Assume that $\mathfrak g = \mathfrak g_1 \oplus \mathfrak g_2$ as Lie algebras (direct sum, i.e. I suppose that they commute). Then $\alpha$ is an isomorphism of algebras, since for every $R$-algebra $A$, the map $$\text{Hom}(\mathfrak g,A) \cong \text{Hom}(U(\mathfrak g),A) \stackrel{\alpha^*}{\to} \text{Hom}(U(\mathfrak g_1) \otimes_R U(\mathfrak g_2),A)$$ is bijective. This is because the right-hand side is given by all $$(f_1,f_2) \in \text{Hom}(U(\mathfrak g_1) \times U(\mathfrak g_2),A) \mbox{ with } [f_1(x_1),f_2(x_2)]=0 \mbox{ for } x_i \in \mathfrak g_i,$$ that is, all pairs $(f_1,f_2)\!: \mathfrak g_1 \oplus \mathfrak g_2 \to A$ with $[f_1(x_1),f_2(x_2)]=0$ for all $x_i \in \mathfrak g_i$. Then the inverse of $\alpha^*$ is defined by $(f_1,f_2) \mapsto f_1+f_2$, which is well-defined, since $$(f_1+f_2)([x_1+x_2,y_1+y_2]) = f_1([x_1,y_1]) + f_2([x_2,y_2]) = [f_1(x_1) + f_2(x_2), f_1(y_1) + f_2(y_2)].$$

I'm adding my comment as a partial answer, as discussed there; this is a reformulation of point 2 that has been added to the question.

Let $\mathcal L$ be the category of Lie $R$-algebras. Assume that $\mathfrak g = \mathfrak g_1 \oplus \mathfrak g_2$ in $\mathcal L$ (direct sum, i.e. I suppose that they commute). Let $\mathcal C$ be the category of unital $R$-algebras. Then $\alpha$ is an isomorphism in $\mathcal C$, since for every unital $R$-algebra $A$, the map $$\text{Hom}_{\mathcal L}(\mathfrak g,A) \cong \text{Hom}_{\mathcal C}(U(\mathfrak g),A) \stackrel{\alpha^*}{\to} \text{Hom}_{\mathcal C}(U(\mathfrak g_1) \otimes_R U(\mathfrak g_2),A)$$ is bijective (by the Yoneda lemma). This is because the right-hand side is given by all $$(f_1,f_2) \in \text{Hom}_{\mathcal C}(U(\mathfrak g_1),A) \times \text{Hom}_{\mathcal C}(U(\mathfrak g_2),A) \mbox{ with } [f_1(x_1),f_2(x_2)]=0 \mbox{ for } x_i \in \mathfrak g_i,$$ that is, all pairs of maps $(f_1,f_2)\!: \mathfrak g_1 \oplus \mathfrak g_2 \to A$ in $\mathcal L$ with $[f_1(x_1),f_2(x_2)]=0$ for all $x_i \in \mathfrak g_i$. Then the inverse of $\alpha^*$ is defined by $(f_1,f_2) \mapsto f_1+f_2$, which is well-defined, since $$(f_1+f_2)([x_1+x_2,y_1+y_2]) = f_1([x_1,y_1]) + f_2([x_2,y_2]) = [f_1(x_1) + f_2(x_2), f_1(y_1) + f_2(y_2)].$$

I'm adding my comment as an answer, as discussed there; this is a reformulation of point 2 that has been added to the question.

Assume that $\mathfrak g = \mathfrak g_1 \oplus \mathfrak g_2$ as Lie algebras (direct sum, i.e. I suppose that they commute). Then $\alpha$ is an isomorphism of algebras, since for every $R$-algebra $A$, the map $$\text{Hom}(\mathfrak g,A) \cong \text{Hom}(U(\mathfrak g),A) \stackrel{\alpha^*}{\to} \text{Hom}(U(\mathfrak g_1) \otimes_R U(\mathfrak g_2),A)$$ is bijective. This is because the right-hand side is given by all $$(f_1,f_2) \in \text{Hom}(U(\mathfrak g_1) \times U(\mathfrak g_2),A) \mbox{ with } [f_1(x_1),f_2(x_2)]=0 \mbox{ for } x_i \in \mathfrak g_i,$$ that is, all pairs $(f_1,f_2)\!: \mathfrak g_1 \oplus \mathfrak g_2 \to A$ with $[f_1(x_1),f_2(x_2)]=0$ for all $x_i \in \mathfrak g_i$. Then the inverse of $\alpha^*$ is defined by $(f_1,f_2) \mapsto f_1+f_2$, which is well-defined, since $$(f_1+f_2)([x_1+x_2,y_1+y_2]) = f_1([x_1,y_1]) + f_2([x_2,y_2]) = [f_1(x_1) + f_2(x_2), f_1(y_1) + f_2(y_2)].$$

I'm adding my comment as a partial answer, as discussed there; this is a reformulation of point 2 that has been added to the question.

Assume that $\mathfrak g = \mathfrak g_1 \oplus \mathfrak g_2$ as Lie algebras (direct sum, i.e. I suppose that they commute). Then $\alpha$ is an isomorphism of algebras, since for every $R$-algebra $A$, the map $$\text{Hom}(\mathfrak g,A) \cong \text{Hom}(U(\mathfrak g),A) \stackrel{\alpha^*}{\to} \text{Hom}(U(\mathfrak g_1) \otimes_R U(\mathfrak g_2),A)$$ is bijective. This is because the right-hand side is given by all $$(f_1,f_2) \in \text{Hom}(U(\mathfrak g_1) \times U(\mathfrak g_2),A) \mbox{ with } [f_1(x_1),f_2(x_2)]=0 \mbox{ for } x_i \in \mathfrak g_i,$$ that is, all pairs $(f_1,f_2)\!: \mathfrak g_1 \oplus \mathfrak g_2 \to A$ with $[f_1(x_1),f_2(x_2)]=0$ for all $x_i \in \mathfrak g_i$. Then the inverse of $\alpha^*$ is defined by $(f_1,f_2) \mapsto f_1+f_2$, which is well-defined, since $$(f_1+f_2)([x_1+x_2,y_1+y_2]) = f_1([x_1,y_1]) + f_2([x_2,y_2]) = [f_1(x_1) + f_2(x_2), f_1(y_1) + f_2(y_2)].$$

I'm adding my comment as an answer, as discussed there; this is a reformulation of point 2 that has been added to the question.

Assume that $\mathfrak g = \mathfrak g_1 \oplus \mathfrak g_2$ as Lie algebras. Then $\alpha$ is an isomorphism of algebras, since for every $R$-algebra $A$, the map $$\text{Hom}(\mathfrak g,A) \cong \text{Hom}(U(\mathfrak g),A) \stackrel{\alpha^*}{\to} \text{Hom}(U(\mathfrak g_1) \otimes_R U(\mathfrak g_2),A)$$ is bijective. This is because the right-hand side is given by all $$(f_1,f_2) \in \text{Hom}(U(\mathfrak g_1) \times U(\mathfrak g_2),A) \mbox{ with } [f_1(x_1),f_2(x_2)]=0 \mbox{ for } x_i \in \mathfrak g_i,$$ that is, all pairs $(f_1,f_2)\!: \mathfrak g_1 \oplus \mathfrak g_2 \to A$ with $[f_1(x_1),f_2(x_2)]=0$ for all $x_i \in \mathfrak g_i$. Then the inverse of $\alpha^*$ is defined by $(f_1,f_2) \mapsto f_1+f_2$, which is well-defined, since $$(f_1+f_2)([x_1+x_2,y_1+y_2]) = f_1([x_1,y_1]) + f_2([x_2,y_2]) = [f_1(x_1) + f_2(x_2), f_1(y_1) + f_2(y_2)].$$

I'm adding my comment as an answer, as discussed there; this is a reformulation of point 2 that has been added to the question.

Assume that $\mathfrak g = \mathfrak g_1 \oplus \mathfrak g_2$ as Lie algebras (direct sum, i.e. I suppose that they commute). Then $\alpha$ is an isomorphism of algebras, since for every $R$-algebra $A$, the map $$\text{Hom}(\mathfrak g,A) \cong \text{Hom}(U(\mathfrak g),A) \stackrel{\alpha^*}{\to} \text{Hom}(U(\mathfrak g_1) \otimes_R U(\mathfrak g_2),A)$$ is bijective. This is because the right-hand side is given by all $$(f_1,f_2) \in \text{Hom}(U(\mathfrak g_1) \times U(\mathfrak g_2),A) \mbox{ with } [f_1(x_1),f_2(x_2)]=0 \mbox{ for } x_i \in \mathfrak g_i,$$ that is, all pairs $(f_1,f_2)\!: \mathfrak g_1 \oplus \mathfrak g_2 \to A$ with $[f_1(x_1),f_2(x_2)]=0$ for all $x_i \in \mathfrak g_i$. Then the inverse of $\alpha^*$ is defined by $(f_1,f_2) \mapsto f_1+f_2$, which is well-defined, since $$(f_1+f_2)([x_1+x_2,y_1+y_2]) = f_1([x_1,y_1]) + f_2([x_2,y_2]) = [f_1(x_1) + f_2(x_2), f_1(y_1) + f_2(y_2)].$$