I'll try to keep your notation except for three things: I prefer to have all parameters non-negative not to get confused myself, I'd rather have $z=(x,y)\in\mathbb R^2$ coordinates on the plane to avoid typing too many subscripts, and I'll replace $L^r$ by $L^q$ because $r$ is too handy for other things like the distance to the origin, etc., so we'll be interested in the inequalities
$$
\| |z|^{-\gamma}u\|_{L^p}\le C(\| |y|^\alpha u_x\|_{L^2}+\| |x|^\alpha u_y\|_{L^2})^a\||z|^\beta u\|_{L^q}^{1-a}\,.
$$
Note that we can split the plane into $4$ coordinate quadrants and deal with each quadrant separately, so I'll restrict the proof to the first quadrant.
Our first task will be to establish the one-dimensional inequality
$$
\int_0^\infty r^{2\alpha-1}|v(r)|^2\,dr\le \frac 1{\alpha^2}\int_0^\infty r^{2\alpha+1}|v'(r)|^2\,dr
$$
for smooth compactly supported $v$ on $[0,+\infty)$ (we do not require $v(0)$ to be $0$, but we require $v$ to vanish outside a bounded interval).
This is equivalent to the $1/\alpha$ norm bound for the linear integral operator with the kernel $K(r,\rho)r^{\alpha-\frac 12}\rho^{-\alpha-\frac 12}\chi_{\{\rho>r\}}$ in $L^2([0,+\infty),dr)$, which follows from the Shur test with $\varphi(t)=\psi(t)=t^{-1/2}$ on $[0,\infty)$.
Applying this to the function $v(r)=u(re^{i\theta})$ with fixed $\theta\in [\frac \pi 9, \frac{4\pi}9]$, say, we conclude that
$$
\int_0^\infty |r^\alpha u(re^{i\theta})/r|^2\,r\,dr\le \frac{1}{\alpha^2}\int_0^\infty |r^\alpha \nabla u(re^{i\theta})|^2\,r\,dr\,.
$$
However, in the angle $A=\{re^{i\theta}:r>0, \theta\in [\frac \pi 9, \frac{4\pi}9]\}$, we have $r^\alpha|\nabla u|$ comparable to $x^\alpha|u_y|+y^\alpha |u_x|$, so we see that we can integrate with respect to $\theta$ now and, switching back from polar to Cartesian, add $\||z|^{-1}u\|_{ L^2( A ) }$ to the first factor on the RHS of our desired inequality without changing anything.
Now we can localize to $D_r=\{z: x,y\le 2r, \max(x,y)\ge r\}=[0,r]\times [r,2r]\cup [r,2r]\times [0,r]\cup[r,2r]\times [r,2r]=Q_1\cup Q_2\cup Q_3$.
We shall first get our estimates in the case $r=1$. What we shall be interested in at this moment is finding the range of $s>0$ for which
$$
\| u\|^2_{L^s(D_1)}\le C\left[\int_{D_1} (|x^\alpha u_y|^2+|y^{\alpha}u_x|^2)+\int_{Q_3}|u|^2\right]
$$
(note that $Q_3\subset A$).
First of all, using the standard 1-dimensional inequality $\int_0^2 |v(x)|^2\,dx \le C\left(\int_0^2 |v'(x)|\,dx+\int_1^2 |v(x)|^2\,dx\right)$ with $v(x)=u(x,y)$, $1\le y\le 2$, we see that the quantity on the right dominates the integral $\int_{Q_1}|u|^2$. Similarly, considering the vertical lines, we can estimate $\int_{Q_2}|u|^2$. Thus we can replace the integral of $|u|^2$ over $Q_3$ on the RHS by that over $D_1$.
Observe next that when $\min(x,y)$ is separated from $0$, the first integral is essentially the same as that of $|\nabla u|^2$, so in that region we can use the classical Sobolev's embedding to conclude that any $s>0$ would work for that part. Thus, we need to take care of the regions near the axes only.
Let now $x\in[0,1], y\in[1,2]$. For every $c\in\mathbb R$ such that the curve $\Gamma_c=\{(x+\xi, y+c\xi^{\alpha+1}):0\le\xi\le 1\}$ is contained in $D_1$ and every $t\in[\frac 12,1]$, we can write
$$
|u(x,y)|=\left|u(x+t, y+ct^{\alpha+1})-\int_0^t \frac d{d\xi}u(x+\xi,y+c\xi^{\alpha+1})\,d\xi\right|
\\
\le \int_0^1[|u_x|+(\alpha+1)x^\alpha |u_y|](x+\xi,y+c\xi^{\alpha+1})\,d\xi
+|u(x+t,y+ct^{\alpha+1})|\,.
$$
Integrating over $c\in[1-y,2-y], t\in[\frac 12,1]$ with respect to $dt\,dc$ and using that $d\xi\, d(c\xi^{\alpha+1})=\xi^{\alpha+1}\,d\xi\,dc$, we conclude that in $Q_1$ we have
$$
|u|\le C\left([(|u_x|+x^{\alpha}|u_y|)\chi_{D_1}]*K+\int_{D_1}|u|\right)
$$
where $K(\xi,\eta)=\frac 1{|\xi|^{\alpha+1}}\chi_{\{-1<\xi<0, |\eta|\le |\xi|^{\alpha+1}\}}$.
The integral term is a constant controlled by $\left(\int_{D_1}|u|^2\right)^{1/2}$, so we just need to figure out for which $s$ the convolution acts from $L^2(D_1)$ to $L^s(D_1)$.
Let $0\le g\in L^2(D_1)$, $\|g\|_{L^2(D_1)}=1$. Let $T>1$. We will estimate the measure of the subset of $D_1$ on which $g*K>T$ . To this end, we'll take $\delta\in(0,1)$ and split
$$
K=K\chi_{\{-\delta<\xi<0\}}+K\chi_{\{\xi<-\delta\}}=K_\delta+K^\delta\,.
$$
Note that
$$
\int |K^\delta|^2=\int_{-1}^{-\delta}\frac 2{|\xi|^{\alpha+1}}\,d\xi\le \frac 2\alpha \delta^{-\alpha}\,
$$
so, choosing $\delta=\left(\frac 8\alpha T^{-2}\right)^{1/\alpha}$ and using Cauchy-Schwarz, we conclude that $\|g*K^{\delta}\|_{L^\infty}\le\frac T2$. On the other hand, $\|K_\delta\|_{L^1}=2\delta$, so $\|g*K_\delta\|_{L^2}\le 2\delta$ and the measure of the set where $g*K_\delta>\frac T2$ is at most $4\delta^2/(T/2)^2\le C T^{-2-\frac 4{\alpha}}$, which means that the convolution operator in question acts from $L^2(D_1)$ to $L^s(D_1)$ at least for $s<2+\frac 4\alpha$. With some more effort, we can get the endpoint result $s=2+\frac 4\alpha$ too but it won't matter because there will be another, stronger, restriction on $s$ coming from your desire to keep $\beta,\gamma\ge 0$.
The upshot so far is that we have the inequality
$$
\int_{D_1} |u|^s\le C\left[\int_{D_1}(|x^\alpha u_y|^2+|y^\alpha u_x|^2)+\int_{D_1\cap A}||z|^{\alpha}(u/|z|)|^2\right]^{s/2}
$$
(I introduced the extra power of $|z|$ on $u$ to have the same scaling with respect to dilations on both 3 terms; in $D_1$ that factor is almost constant). Now, plugging in $U(z)=u(rz)$ instead of $U$ and making the change of variables carefully, we arrive at
$$
r^{-2}\int_{D_r}|u|^s\le C\left(r^{-2\alpha}\left[\int_{D_r}(|x^\alpha u_y|^2+|y^\alpha u_x|^2)+\int_{D_r\cap A}||z|^{\alpha}(u/|z|)|^2\right]\right)^{s/2}\,,
$$
i.e.,
$$
\int_{D_r}|u|^s\le r^{2-\alpha s} J_r^{s/2}
$$
where $\sum_{r=2^k; k\in\mathbb Z} J_r\le C\|x^\alpha |u_y|+y^{\alpha}|u_x|\|_{L^2}^2$,.
If we now choose any $\tau\in[0,1]$, we can use Holder to write
$$
\int_{D_r} |u|^{\tau s+(1-\tau) q}\le \left[\int_{D_r}|u|^s\right]^\tau\left[\int_{D_r}|u|^q\right]^{1-\tau}\le r^{(2-\alpha s)\tau}J_r^{\tau s/2}\left[\int_{D_r}|u|^q\right]^{1-\tau}\,.
$$
Now set $p=\tau s+(1-\tau) q$ and choose $\beta,\gamma\ge 0$ so that
$$
\gamma p+\beta q(1-\tau)=(2-\alpha s)\tau
$$
(which will require $2-\alpha s\ge 0$, i. e., a stronger restriction than we had before). Then we can rewrite this as
$$
\int_{D_r} ||z|^{-\gamma}u|^p\le J_r^{\tau s/2}\left(\int_{D_r} ||z|^\beta u|^q \right)^{1-\tau}
$$
whence, adding these inequalities up and using
$$
\sum_r J_r^{\tau s/2} I_r^{1-\tau}\le\left(\sum_r J_r^{s/2}\right)^\tau
\left(\sum_r I_r\right)^{1-\tau}\le \left(\sum_r J_r\right)^{\tau s/2}
\left(\sum_r I_r\right)^{1-\tau}
$$
for $s\ge 2$, we finally obtain
$$
\||z|^{-\gamma} u\|_{L^p}\le \|x^\alpha |u_y|+y^\alpha |u_x|\|_{L^2}^a\||z|^\beta u\|_{L^q}^{1-a}
$$
with $a=\tau s/p$.
The last thing is to comb everything up to get some decent form of our parametric representations. We have
$$
p=\tau s+(1-\tau) q
\\
\gamma p+\beta q(1-\tau)=(2-\alpha s)\tau
\\
a=\tau s/p
\\
\tau\in[0,1]
\\
s\in [2,\tfrac 2\alpha]
$$
which, after some algebra, finally turns into (unless I've made a stupid error)
$$
\frac {a\alpha}2\le \frac 1p-\frac{1-a}q\le \frac a2
\\
\gamma+(1-a)\beta=2(\tfrac 1p-\tfrac{1-a}q)-a\alpha\,.
\\
p,q>0, a\in[0,1], \gamma,\beta\ge 0
$$
Whether this range is sufficient for your purposes is for you to figure out. If yes, great. If no, just post what values you'd like to have (though I don't think the range can be extended, so in that case I'll try to construct a counterexample).
