First, lets say we have some variable $y$, then by the sifting property of delta functions, can we say the following is true?
$$ \int \exp(\frac{d}{d x}) \delta(x-y)dx f(y) = \exp(\frac{d}{d y})f(y)~? \tag{1}$$
where $\delta(\cdot)$ is the Dirac delta function.
If thats true, then what if we replace $y$ in the delta function with a constant (say $0$). The derivative with respect to a constant is meaningless, does that mean it would be 0? That is, does the following make sense?
$$ \int \exp(\frac{d}{d x}) \delta(x)dxf(y) = \exp(\frac{d}{d 0})f(y) = f(y)~? \tag{2} $$
$T_1=\exp(\frac{d}{d x})$ is the translation operator $(T_1f)(x)=f(x+1)$. So OP's expressions become $$\int_{\mathbb{R}} \exp(\frac{d}{d x}) \delta(x-y)dx f(y)~=~\int_{\mathbb{R}} \delta(x+1-y)dx f(y) ~=~f(y) \tag{1}$$ and $$\int_{\mathbb{R}} \exp(\frac{d}{d x}) \delta(x)dx f(y)~=~\int_{\mathbb{R}} \delta(x+1)dx f(y) ~=~f(y). \tag{2}$$