Extending partial sums of the Taylor series of $e^x$ to a smooth function on $\mathbb{R}^2$?

Sure; we have the relationship

$$\sum_{k=0}^n \frac{x^k}{k!}=\frac{\Gamma(n+1,x)}{\Gamma(n+1)}\exp\,x=Q(n+1,x)\exp\,x$$

where $\Gamma(n)$ is the usual gamma function and $\Gamma(n,a)$ is the (upper) incomplete gamma function, and $Q(n,a)$ is the regularized (upper) incomplete gamma function; see that link as well as this link for more on its properties. The expression on the right definitely makes sense not only for integer values of $n$...

This is meant as a compliment to J.M.'s answer but the interpolation provided has a connection to fractional derivatives. Specifically, $$ (1-D_x^{-(\nu+1)})e^x = e^x Q(\nu+1,x), $$ where $D_x^\alpha$ is interpreted to be the Riemann-Liouville fractional derivative w.r.t. the variable $x$. This is easily seen by first expanding $e^x$ as a power series to write $$ (1-D_x^{-(\nu+1)})e^x% =e^x-\sum_{k=0}^\infty D_x^{-(\nu+1)}\frac{x^k}{k!}% =e^x-\sum_{k=0}^\infty \frac{\Gamma(k+1)}{\Gamma(\nu+k+2)}\frac{x^{k+\nu+1}}{k!}. $$ Then, with a bit of algebraic manipulation we obtain $$ (1-D_x^{-(\nu+1)})e^x% =e^x-\frac{z^{\nu+1}}{\Gamma(\nu+2)}\sum_{k=0}^\infty \frac{(1)_k}{(\nu+2)_k}\frac{x^k}{k!}% =e^x-\frac{z^{\nu+1}}{\Gamma(\nu+2)}{_1F_1}(1,\nu+2,x)=e^x-e^x P(\nu+1,x)=e^xQ(\nu+1,x), $$ where $(a)_k=\Gamma(a+k)/\Gamma(a)$ is the Pochhammer symbol and $P(z,s)$ is the regularized (lower) incomplete gamma function.

EDIT: So I did a little more digging and came across the article Expansion of fractional derivatives in terms of an integer derivative series

Equation $(15)$ for $a=0$ gives the following integer derivative expansion of the Riemann-Liouville fractional derivative $$ D_x^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}\frac{x^{k-\alpha}}{\Gamma(k-\alpha+1)}D_x^k. $$ If we treat this expansion as a traditional power series we find $$ D_x^\alpha=x^{-\alpha}{_1\mathbf F_1}(-\alpha,1-\alpha,-xD_x), $$ where ${_1\mathbf F_1}(a;b;z)=\frac{1}{\Gamma(b)}{_1F_1}(a;b;z)$ is the regularized confluent hypergeometric function of the first kind. Since $D_x^k \,e^x=e^x\ \forall\, k\in\Bbb N_0$ the derivative is evaluated to be $$ \begin{aligned} (1-D_x^{-(\nu+1)})e^x% &=e^x-x^{-\alpha}{_1\mathbf F_1}(\nu+1,\nu+2,-xD_x)e^x\\ &=e^x-x^{\nu+1}{_1\mathbf F_1}(\nu+1,\nu+2,-x)e^x\\ &=e^x(1-P(\nu+1,x))\\ &=e^xQ(\nu+1,x). \end{aligned} $$