The Riemann-Liouville formula for the semiderivative goes like this:
${}_0 D_x^{(1/2)} f(x)=\frac{f(0)}{\sqrt{\pi x}}+\frac1{\sqrt\pi}\int_0^x \frac{f^\prime(t)}{\sqrt{x-t}}\mathrm dt$
Taking $f(x)=\exp(-\alpha x^2+\beta x)$, we then have
$\begin{align*} {}_0 D_x^{(1/2)} f(x)&=\frac1{\sqrt{\pi x}}+\frac1{\sqrt\pi}\int_0^x \frac{(\beta-2\alpha t)\exp(-\alpha t^2+\beta t)}{\sqrt{x-t}}\mathrm dt\\ &=\frac1{\sqrt{\pi x}}+\frac1{\sqrt\pi}\left(\beta\int_0^x \frac{\exp(-\alpha t^2+\beta t)}{\sqrt{x-t}}\mathrm dt-2\alpha\int_0^x \frac{t\exp(-\alpha t^2+\beta t)}{\sqrt{x-t}}\mathrm dt\right) \end{align*}$
Unfortunately, neither integral seems to have an easy evaluation in terms of known functions unless $\alpha=0$ or $\beta=0$, in which case,
$\begin{align*} {}_0 D_x^{(1/2)} \exp(-\alpha x^2)&=\frac1{\sqrt{\pi x}}-\frac{8\alpha x^{3/2}}{3\sqrt\pi}{}_2 F_2\left({{1,\frac32}\atop{\frac54,\frac74}}\mid -\alpha x^2\right)\\ {}_0 D_x^{(1/2)} \exp(\beta x)&=\frac1{\sqrt{\pi x}}+\sqrt{\beta}\exp(\beta x)\mathrm{erf}(\sqrt{\beta x}) \end{align*}$
There is in fact a semiderivative version of the Leibniz formula, but as I said, it looks somewhat unwieldy:
${}_0 D_x^{(1/2)}f(x)g(x)=\sum_{j=0}^\infty \binom{\frac12}{j}\left({}_0 D_x^{(1/2-j)}f(x)\right)g^{(j)}(x)$
If we take $f(x)=\exp(\beta x)$ and $g(x)=\exp(-\alpha x^2)$, we get
$\begin{split}&{}_0 D_x^{(1/2)} \exp(-\alpha x^2+\beta x)=\\&\quad\exp(-\alpha x^2+\beta x)\sum_{j=0}^\infty \binom{\frac12}{j}\frac{(-1)^j \alpha^{j/2} \beta^{1/2-j}}{\Gamma(j-1/2)}\gamma(j-1/2,\beta x) H_j(x\sqrt\alpha)\end{split}$
where $\gamma(a,x)$ is a (lower) incomplete gamma function, and $H_j(x)$ is a Hermite polynomial. I know of no simpler way to express this infinite sum, however.