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$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=\frac{{a_1}}{x} \left(\left({\lambda}+x^2\right) M(-\frac{{\lambda}}{2},\frac{1}{2},x^2)-({\lambda}-2) M(1-\frac{{\lambda}}{2},\frac{1}{2},x^2)\right)+2 {a_2} {\lambda} H({\lambda}-1,x)$$

for $\lambda$ negative non-integer and $x\rightarrow-\infty$ the $a_1$ term diverges as $e^{x^2/2}$, while the $a_2$ term diverges as $e^{x^2}$, so it does not seem possible to cancel the divergences by a suitable choice of $a_1$, $a_2$.

$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=-2a_1\lambda x M(1-\tfrac{1}{2}{\lambda},\tfrac{3}{2},x^2)+2 {a_2} {\lambda} H({\lambda}-1,x)$$ The asymptotics for $x\rightarrow-\infty$ and $\lambda$ negative non-integer is $$y'(x)\rightarrow\frac{2 e^{x^2}(-x)^{-\lambda} }{\sqrt{\pi }\, \Gamma \left(-\lambda/2\right)}\bigl(a_2 \Gamma \left(-\lambda/2\right) \Gamma (\lambda+1)\sin \pi \lambda -\pi a_1\bigr).$$ So this vanishes if $$a_2 \Gamma \left(-\lambda/2\right) \Gamma (\lambda+1)\sin \pi \lambda =\pi a_1.$$

$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=\frac{{a_1}}{x} \left(\left({\lambda}+x^2\right) M(-\frac{{\lambda}}{2},\frac{1}{2},x^2)-({\lambda}-2) M(1-\frac{{\lambda}}{2},\frac{1}{2},x^2)\right)+2 {a_2} {\lambda} H({\lambda}-1,x)$$

for $\lambda$ negative non-integer and $x\rightarrow-\infty$ the $a_1$ term diverges as $e^{x^2/2}$, while the $a_2$ diverges as $e^{x^2}$, so it does not seem possible to cancel the divergence by a suitable choice of $a_1$, $a_2$.

$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=\frac{{a_1}}{x} \left(\left({\lambda}+x^2\right) M(-\frac{{\lambda}}{2},\frac{1}{2},x^2)-({\lambda}-2) M(1-\frac{{\lambda}}{2},\frac{1}{2},x^2)\right)+2 {a_2} {\lambda} H({\lambda}-1,x)$$

for $\lambda$ negative non-integer and $x\rightarrow-\infty$ the $a_1$ term diverges as $e^{x^2/2}$, while the $a_2$ term diverges as $e^{x^2}$, so it does not seem possible to cancel the divergences by a suitable choice of $a_1$, $a_2$.

if I'm not mistaken $\partial H(\lambda,x)/\partial x\rightarrow 2^{\lambda} \lambda x^{\lambda-1}$ for $x\rightarrow \infty$, so the $a_2$ term vanishes for any $\lambda<1$; hence if the condition you are looking for exists for a nonzero $a_2$, it can only involve $\lambda$, and then the condition seems to be $\lambda=-2,-4,-6,\ldots$ (for any other value of $\lambda$ the derivative blows up $\propto e^{x^2/2}$).

$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=\frac{{a_1}}{x} \left(\left({\lambda}+x^2\right) M(-\frac{{\lambda}}{2},\frac{1}{2},x^2)-({\lambda}-2) M(1-\frac{{\lambda}}{2},\frac{1}{2},x^2)\right)+2 {a_2} {\lambda} H({\lambda}-1,x)$$

for $\lambda$ negative non-integer and $x\rightarrow-\infty$ the $a_1$ term diverges as $e^{x^2/2}$, while the $a_2$ diverges as $e^{x^2}$, so it does not seem possible to cancel the divergence by a suitable choice of $a_1$, $a_2$.