Density of a linear transformation of an inverse gaussian variable

I did find a general answer to my problem.

Suppose that you have two random variables, $X, Y: \Omega \rightarrow \mathbb{R}$ and two non random parameters $(a,b) \in \mathbb{R}$ related by $Y = aX + b$. Furthermore, let $f_X$ and $f_Y$ be their respective density functions. Then,

\begin{align} f_Y(y) = \frac{1}{|a|} f_X(x). \end{align}

The proof is actually fairly simple. I'll do it for $a > 0$:

\begin{align} \mathbb{P}(u < Y < v) &= \mathbb{P}\left( u < aX + b < v \right) \\ &= \mathbb{P}\left( \frac{u-b}{a} < X < \frac{v-b}{a} \right) \\ &= \int_{\frac{v-b}{a}}^{\frac{u-b}{a}} f_X(x) dx \end{align}

Then, we proceed with a change of variable: $y = ax + b$ implies $x = \frac{y-b}{a}$ and $dx = \frac{1}{a}$ such that

\begin{align} \int_{\frac{v-b}{a}}^{\frac{u-b}{a}} f_X(x) dx = \int_{v}^{u} \frac{1}{|a|} f_X\left(\frac{y-b}{a}\right) dy := \int_u^v f_Y(y) dy \end{align}

where we thus have defined the density $f_Y(y) := \frac{1}{|a|} f_X\left(\frac{y-b}{a}\right) = \frac{1}{|a|} f_X(x)$. The case for $a < 0$ follows the same logic, except one negative sign appears when using the survival function and another when using $-|a| = a$, so they cancel out and you recover the same relationship.