Let $R$ a domain. Let $f$ be a monic polynomial with coefficients in $R$ and degree 2 such that there exists an isomorphism $\phi$ between $R[X]/(f)$ and $R \times R$. Suppose that there exist different $a, b$ such that $\phi(X) = (a, b)$. Prove that (1) both $f(a)$ and $f(b)$ are equal to $0$, and (2) that $a - b$ is a unit.
For (1) I did the following. Write $f = X^2 + cX + d$, then $(0, 0) = \phi(0) = \phi(f(X)) = (a^2, b^2) + \phi(c)(a, b) + \phi(d)$. So it be nice if $\phi(c) = (c, c)$ for $c$ in $R$, but I'm not sure if that is true and how to prove that.
For (2) we got the hint to look at $p(X)$ such that $\phi(p(X)) = (1, 0)$. Then prove that $p(a) = 1$ and $p(b) = 0$. I think if I know how to proof (1), this will be analogous. Suppose that this is proven, then $p(a) - p(b) = 1$, and since $p(X) - p(b)$ is zero for $X = b$, we can factor $(X - b)$ out of $p(X) - p(b)$, and putting in $X = a$ gives $1 = p(a) - p(b) = (a - b)c(a)$ for some polynomial $c(a)$, which will be the inverse of $a - b$. Is this correct?
Original answer (without the assumption $a\ne b$)
The statement is false. Consider $R=\mathbb{C}$ and the map $$ \alpha\colon\mathbb{C}[X]\to\mathbb{C}\times\mathbb{C} \qquad \alpha(g)=\bigl(\,\overline{g(i)},g(-i)\bigr) $$ which is readily seen to be a ring homomorphism.
Then $\alpha$ is surjective. Indeed, if $r,s\in\mathbb{C}$, we need to find $g(X)$ so that $g(i)=\bar{r}$ and $g(-i)=s$ and we can look for $g(X)=pX+q$, getting $$ \begin{cases} pi+q=\bar{r}\\ -pi+q=s \end{cases} $$ As $$ \det\begin{bmatrix}i & 1 \\ -i & 1\end{bmatrix}=2i\ne0 $$ the linear system has a unique solution.
Note that $\alpha(g)=(0,0)$ if and only if $g(i)=g(-i)=0$, that is, if and only if $g\in(X^2+1)$. So we can take $f(X)=X^2+1$ and $\alpha$ induces an isomorphism $\phi\colon\mathbb{C}[X]/(f)\to\mathbb{C}\times\mathbb{C}$ such that $$ \phi(X+(f))=\alpha(X)=(\bar{i},-i)=(-i,-i) $$ So in this case $a=-i=b$ and $a-b=0$ is not a unit.
Answer for the case $a\ne b$.
The statement is false also for the case $a\ne b$. Consider the polynomial $f(X)=(X-1-i)(X-i)\in\mathbb{Z}[i][X]$. Then you can see that the homomorphism $$ \alpha\colon\mathbb{Z}[i][X]\to\mathbb{Z}[i]\times\mathbb{Z}[i] $$ given by $$ \alpha(g)=(g(1+i),g(i)) $$ is a surjective homomorphism (because $(1+i)-i$ is invertible). Its kernel is $(f)$. Now compose this with the isomorphism $$ \mathbb{Z}[i]\times\mathbb{Z}[i]\to\mathbb{Z}[i]\times\mathbb{Z}[i] \qquad (x,y)\mapsto(\bar{x},y) $$ and, like before, you'll get an isomorphism $\phi\colon\mathbb{Z}[i][X]\to\mathbb{Z}[i]\times\mathbb{Z}[i]$ such that $$ \phi(X+(f))=(1-i,i) $$ and $(1-i)-i=1-2i$ is not invertible.