No.
Some words follow, to explain how i understood the question, and why i claim the negative answer. An example may make the situation simpler to explain. Let us consider over $F=\Bbb F_p$, $p=11$, the elliptic curve with equation $y^2=x^3 + x + 1$. Then on the curve we have the rational points
$$ P_\pm = (0,\pm 1)\ ,\qquad Q_\pm=(1,\pm 5)\ .$$
Then
- the "two $P$-points" have the same $x$-component, $x_p=0$,
- the "two $Q$-points" have the same $x$-component, $x_q=1$,
but adding points in all four possible ways to combine deliver, here in sage:
sage: E = EllipticCurve( GF(11), [1, 1] )
sage: P0, P1 = E.point( (0, +1) ), E.point( (0, -1) )
sage: Q0, Q1 = E.point( (1, +5) ), E.point( (1, -5) )
sage: [ (P+Q).xy() for P in [P0, P1] for Q in [Q0, Q1] ]
[(4, 5), (2, 0), (2, 0), (4, 6)]
we get the above points, the first coordinate is either $2$, or $4$. So there is no "unique $x_r$" that can be associated only with the knowledge of $x_p$, $x_q$.
The reason can be simply explained by the fact, that the computation of $R$ depends of building the following slope $m$ in the affine plane of the rational points $P(x_p,y_p)$, $Q(x_q,y_q)$,
$$
m = \frac{y_p-y_q}{x_p-x_q}\ .
$$
So exchanging signs in $\pm y_p$, $\pm y_q$ leads "very often" to a difference.
Bonus: Here are some further thoughts related to the geometrical construction of $P+Q$ and $2P:=P+P$ for two points $P,Q$ on some elliptic curve defined over some finite field $F=\Bbb F_q$, $q$ prime power (of a prime not equal to two, three maybe), or over an infinite field $F=\Bbb Q$ or $\Bbb R$ or $\Bbb C$, by an equation of the shape:
$$
E\ :\ y^2 =x^3+ax+b\ ,\ a,b\in F\ ,
$$
where $x^3+ax+b$ has different roots in some algebraic closure of $F$.
We start with two $F$-rational points $P_1(x_1,y_1)$, and $P_2(x_2,y_2)$ in $E(F)$. In the generic case $x_1\ne x_2$ the equation of the line through the two points is of the shape $y=mx+n$, with $m,n$ depending on $P_1$, $P_2$. Then from
$$
\begin{aligned}
y_1 &=mx_1+n\ ,\\
y_2 &=mx_2+n\ ,&&\text{ after subtraction we get}\\
y_1-y_2&=m(x_1-x_2)\ ,&&\text{ so we have the formula for the slope $m$}\\
m &=\frac{y_1-y_2}{x_1-x_2}\ .
\end{aligned}
$$
For the first coordinate $x_3$ of $\pm (P_1+P_2)=(x_3,\pm y_3)$ we need only $m$, but $n$ can also be found easily from the above relations, then the third point on the line $P_1P_2$ (counting multiplicities) is by the definition of the operation $+$ on $E(F)$ the point $-(P_1+P_2)=(x_3,-y_3)$. (If we denote the coordinates in the sum $P_1+P_2$ by $x_3,y_3$.)
The points
$(x_1,y_1)$,
$(x_2,y_2)$,
$(x_3,-y_3)$ satisfy thus both equations $y^2=x^3+ax+b$ and $y=mx+n$. We eliminate $y$ using the second equation from the first one, so
$$
x^3+ax+b-(mx+n)^2=0\ .
$$
Explicitly:
$$
x^3-m^2x^2+\text{(lower degree terms in $x$)}=0
$$
has then the roots $x_1,x_2,x_3$. Vieta gives us for the sum
$$
x_1+x_2+x_3 = m^2\ ,
$$
so we obtain the formula for the first coordinate $x_3$, which is
algebraically
$$
\boxed{\qquad
\begin{aligned}
x_3 &=-x_1-x_2+m^2
\\
&=-x_1-x_2+\left(\frac{y_1-y_2}{x_1-x_2}\right)^2\ .
\end{aligned}
\qquad}
$$
This is a good point to check the formula for some points on some curve. I will work first with $E$ given by $y^2=x^3+x+1$ over $F=\Bbb F_{11}$, ask for the sum of some random points $P,Q$ in $E(F)$.
We ask for the implemented formula, thus getting $P+Q$ from sage, then i will implement with bare hands the above formula, also ask for the result for the same points.
sage: E = EllipticCurve( GF(11), [1, 1] )
sage: import random
sage: P = random.choice(E.points())
sage: Q = random.choice(E.points())
sage: P, Q, P+Q
((2 : 0 : 1), (0 : 10 : 1), (1 : 6 : 1))
sage: def x3(P, Q):
....: """We compute the first component x3 of P=:(x1, y1), Q=:(x2, y2)
....: using the formula x3 = -x1 -x2 + (y1-y2)^2/(x1-x2)^2
....: Note that we assume x1 != x2, and that P, Q != infinit point"""
....: x1, y1 = P.xy()
....: x2, y2 = Q.xy()
....: return -x1 -x2 + (y1-y2)^2/(x1-x2)^2
....:
....:
sage: x3(P, Q)
1
sage: # the abover is the first component of P+Q = (1, 6)
Let us use some "more convincing" framework.
sage: E = EllipticCurve( QQ, [1, 1] )
sage: E.rank()
1
sage: E.gens()
[(0 : 1 : 1)]
sage: G = E.point( (0, 1) ) # the generator
sage: P, Q = 7*G, -5*G
sage: P, Q
((-3596697936/8760772801 : 591456591665497/819999573400799 : 1),
(43992/82369 : 30699397/23639903 : 1))
sage: P+Q
(1/4 : -9/8 : 1)
sage: x3(P, Q)
1/4
sage: P-Q
(516800901506579137034097949153/116165201153061098261023776144
: 382844998133375068925120757216593508841494737/39592604638617085219380314122331748004030272
: 1)
sage: x3(P, -Q)
516800901506579137034097949153/116165201153061098261023776144
A last remark about doubling. We start with "$P$ and $P$", and need to get the corresponding line $y=mx+n$ again. We use the notation $P(x_1,y_1)$, compute $2P=(x_3,y_3)$, but only the first component. (The second one follows, typing is only harder.)
One possibility to proceed is as follows. The point $P(x_1,y_1)$ is on the curve with equation $F(x,y)=0$, where $F(x,y)=x^3+ax+b-y^2$. Let us write $x=x_1+s$, $y=y_1+t$ to get in the Taylor expansion of $F$ around $(x_1,y_1)$ quickly the linear term. Explicitly:
$$
\begin{aligned}
F(x,y)
&=(x_1+s)^3+a(x_1+s)+b-(y_1+t)^2
\\
&=\underbrace{(x_1^3+ax_1+b-y_1^2)}_{=0}
+(3x_1^2s+as-2y_1t)
+\text{(Higher order monomials in $s,t$)}\ .\\[3mm]
\operatorname{Taylor}_1(F)(x,y)
&=
3x_1^2s+as-2y_1t
\\
&=(3x_1^2+a)(x-x_1)-2y_1(y-y_1)\ .\\[3mm]
&\text{Compare with $y=mx+n$. So the needed slope $m$ is}
\\
m &=\frac{3x_1^2+a}{2y_1}\ .\\[3mm]
&\text{Alternatively:}
\\
m &=
\lim\frac{y_2-y_1}{x_2-x_1}
\\
&=\lim\frac{(y_2-y_1)(y_2+y_1)}{(x_2-x_1)(y_2+y_1)}
\\
&=\lim\frac{y_2^2-y_1^2}{(x_2-x_1)(y_2+y_1)}
\\
&=\lim\frac{(x_2^3+ax_2+b)-(x_1^3+ax_1+b)}{(x_2-x_1)(y_2+y_1)}
\\
&=\lim\frac{x_2^2+x_1x_2+x_1^2+a}{y_2+y_1}
\\
&=\frac{x_1^2+x_1x_1+x_1^2+a}{y_1+y_1}
\\
&=\frac{3x_1^2+a}{2y_1}\ .
\end{aligned}
$$
(The limit is taken for $(x_2,y_2)$ converging on the curve to $(x_1,y_1)$.
From here, we have the same, so $x_3=-x_1-x_1+m^2$. Computer check:
sage: def x3_for_double(P):
....: """We compute the first component x3 for 3P for P=:(x1, y1),
....: using the formula x3 = -2x1 + (3*x1^2+a)^2/4/y1^2
....: """
....: x1, y1 = P.xy()
....: a = P.curve().a4()
....: m = (3*x1^2 + a) / 2 / y1
....: return -2*x1 + m^2
....:
sage: twoP.xy()[0] # first component of 2P
23419679382776533016338728874246651427713/12258805697617629893689847027495187248836
sage: x3_for_double(P)
23419679382776533016338728874246651427713/12258805697617629893689847027495187248836
sage: # here, P has the last value used above
sage: P
(-3596697936/8760772801 : 591456591665497/819999573400799 : 1)
Note: All the above is "naive and standard", but i tried to write down explicitly the algebraic geometry computations. Monographs do not have the space for this, courses do not have the time.
Best Answer
Instead of following blindly Wikipedia's formulas, it is best to understand how to calculate $P+Q$, or $P+P$, given an elliptic curve. Let us assume for simplicity that the curve is given by $E:y^2=x^3+Ax+B$, and $P,Q\in E$.
In order to find $P+Q$, first find the equation of the line $L$ through $P$ and $Q$, find the third point $R$ of intersection of $L$ and $E$. Then $P+Q+R=\mathcal{O}$, so that $R=-(P+Q)$. In this case $-(x_0,y_0)=(x_0,-y_0)$, so we can find $P+Q$ as $-R$.
In order to find $P+P$, first find the tangent line $L$ to $E$ at $P$, and find the third point $R$ of intersection of $L$ and $E$. Then, $2P+R=\mathcal{O}$, and so $2P=-R$.
In your case, $E: y^2=x^3+1$ and $P=(0,1)$. The tangent line is $y=1$, and it turns out that this line has a triple point of intersection with $E$, thus $R=P$. In particular, $2P=-R=-P$, and so $3P=\mathcal{O}$.
Another example with the same curve $E$: let $P=(2,3)$ and $Q=(0,1)$. The line through $P$ and $Q$ is $y=x+1$, and the third point of intersection is $R=(-1,0)$. Since $R=-R$, we obtain $P+Q=(-1,0)$.
Finally, here is a picture of $P=(2,3)$, $2P$, $3P$, $4P$, and $5P$, which shows that $6P=\mathcal{O}$.