$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$This is true. Let $a \in A$. Then $|a|$ also belongs to $A$, and since $A$ is nonunital a little bit of functional calculus shows that there exists $x \in A$ with $\norm x = 1$ and $\norm{\abs a x}$ arbitrarily small. (I guess you have to consider the cases where $0$ is or is not isolated in the spectrum of $\abs a$ separately, unless there's an easier argument I'm missing.) Supposing $A$ is sitting inside $B(H)$, we can write $a = u\abs a$ where $u \in B(H)$ is a partial isometry, and then calculate $$\norm{(I + a)x} = \norm{(I + u\abs a)x} \geq \norm x - \norm{u\abs a x} = 1 - \norm{\abs x}.$$ Since $\norm{\abs a x}$ can be made arbitrarily small, this shows that the norm of $I + a$ in $B(A)$ is at least $1$.

I'm sure this is something many people know, but I don't have any reference.

$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$This is true. Let $a \in A$. Then $|a|$ also belongs to $A$, and since $A$ is nonunital a little bit of functional calculus shows that there exists $x \in A$ with $\norm x = 1$ and $\norm{\abs a x}$ arbitrarily small. (I guess you have to consider the cases where $0$ is or is not isolated in the spectrum of $\abs a$ separately, unless there's an easier argument I'm missing.) Supposing $A$ is sitting inside $B(H)$, we can write $a = u\abs a$ where $u \in B(H)$ is a partial isometry, and then calculate $$\norm{(I + a)x} = \norm{(I + u\abs a)x} \geq \norm x - \norm{u\abs a x} = 1 - \norm{\abs a x}.$$ Since $\norm{\abs a x}$ can be made arbitrarily small, this shows that the norm of $I + a$ in $B(A)$ is at least $1$.

I'm sure this is something many people know, but I don't have any reference.

This is true. Let $a \in A$. Then $|a|$ also belongs to $A$, and since $A$ is nonunital a little bit of functional calculus shows that there exists $x \in A$ with $\|x\| = 1$ and $\|\,|a|x\|$ arbitrarily small. (I guess you have to consider the cases where $0$ is or is not isolated in the spectrum of $|a|$ separately, unless there's an easier argument I'm missing.) Supposing $A$ is sitting inside $B(H)$, we can write $a = u|a|$ where $u \in B(H)$ is a partial isometry, and then calculate $$\|(I + a)x\| = \|(I + u|a|)x\| \geq \|x\| - \|u|a|x\| = 1 - \|\,|a|x\|.$$ Since $\|\, |a|x\|$ can be made arbitrarily small, this shows that the norm of $I + a$ in $B(A)$ is at least $1$.

I'm sure this is something many people know, but I don't have any reference.

$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$This is true. Let $a \in A$. Then $|a|$ also belongs to $A$, and since $A$ is nonunital a little bit of functional calculus shows that there exists $x \in A$ with $\norm x = 1$ and $\norm{\abs a x}$ arbitrarily small. (I guess you have to consider the cases where $0$ is or is not isolated in the spectrum of $\abs a$ separately, unless there's an easier argument I'm missing.) Supposing $A$ is sitting inside $B(H)$, we can write $a = u\abs a$ where $u \in B(H)$ is a partial isometry, and then calculate $$\norm{(I + a)x} = \norm{(I + u\abs a)x} \geq \norm x - \norm{u\abs a x} = 1 - \norm{\abs x}.$$ Since $\norm{\abs a x}$ can be made arbitrarily small, this shows that the norm of $I + a$ in $B(A)$ is at least $1$.

I'm sure this is something many people know, but I don't have any reference.