Yes, they are the same.

Let us take the linear convolution of $x[t]$ and $y[t]$ as 2 portions $H_1$ of length $N$ corresponding to first $N$ samples, and $H_2$ of remaining $N-1$ samples. Circular convolution between $x$ and $y$ causes $H_2$ to overlap over $H_1$ because of time-aliasing. So the first $N-1$ samples of the result is $H_1 + H_2$ with only the last sample being correct (corresponding to linear convolution).

Now, for the linear convolution between $\tilde{x}$ and $\tilde{y}$, the samples from $N$ to $2N-1$ will have additional component of first $N$ samples of $conv(x,y)$. So these samples will correspond circular convolution of $x$ and $y$.

Yes, they are the same.

Let us take the linear convolution of $x[t]$ and $y[t]$ as 2 portions $H_1$ of length $N$ corresponding to first $N$ samples, and $H_2$ of remaining $N-1$ samples. Circular convolution between $x$ and $y$ causes $H_2$ to overlap over $H_1$ because of time-aliasing. So the first $N-1$ samples of the result is $H_1 + H_2$ with only the last sample being correct (corresponding to linear convolution).

Now, for the linear convolution between $\tilde{x}$ and $\tilde{y}$, the samples from $N$ to $2N-1$ will have additional component of first $N$ samples of $\textrm{conv}(x,y)$. So these samples will correspond circular convolution of $x$ and $y$.