The answer is in fact yes. First, construct the following figure consisting of four $2 \times 3$ rectangles. [asy][asy] import geometry; size(10cm); int W=5; int H=6; for (int i=-W; i<=W; ++i) { draw( (i,-H+1)--(i,H), rgb("333333")+dotted ); } for (int i=-H+1; i<=H; ++i) { draw( (-W,i)--(W,i), rgb("333333")+dotted ); } pen purpleL = purple+white+white; pen pinkL = pink+white+white; void drawL(real a, real b, pen c=pinkL, pen d=pinkL) { filldraw(shift(0,0.5+b*-0.5)*((2*a,2*b)--(4*a,2*b)--(4*a,4*b)--(3*a,4*b)--(3*a,3*b)--(2*a,3*b)--cycle), c); filldraw(shift(0,0.5+b*-0.5)*((2*a,3*b)--(2*a,5*b)--(4*a,5*b)--(4*a,4*b)--(3*a,4*b)--(3*a,3*b)--cycle), d); } drawL(1,1); drawL(-1,1, purpleL); drawL(1,-1); drawL(-1,-1); draw(line((0,0), (1,1)), dashed+red); draw(line((0,0.5), (1, 0.5)), dashed+red+blue); draw(line((0,0), (0, 1)), dashed+red+blue+blue); [/asy][/asy] This can be done by reflecting about the red, pink, purple lines in that order starting from the purple $L$-tromino. For simplicity, we can then take a scaling that instead makes it so we are considering $1 \times 1$ squares that are placed as follows. [asy][asy] import geometry; size(5cm); int W=3; int H=3; for (int i=-W; i<=W; ++i) { draw( (i,-H+1)--(i,H), rgb("333333")+dotted ); } for (int i=-H+1; i<=H; ++i) { draw( (-W,i)--(W,i), rgb("333333")+dotted ); } pen pinkL = pink+white+white; filldraw(shift(1,1)*unitsquare, pinkL); filldraw(shift(1,-1)*unitsquare, pinkL); filldraw(shift(-2,-1)*unitsquare, pinkL); filldraw(shift(-2,1)*unitsquare, pinkL); [/asy][/asy] Afterwords, we can then solve covering the plane vertically and horizontally separately now by reflecting along the red lines drawn and repeating. Claim: In the one dimensional case of covering $\mathbb{Z}$ and reflecting across some real $r + x \to r - x$, both the tuples $(1, 0, 1)$ and $(1, 0, 0, 1)$ tile where the tuples represent consecutive elements of $\mathbb{Z}$. Proof. The $(1, 0, 1)$ follows immediately by reflecting to get $(1, 1, 1, 1)$, and then reflecting to get a tuple $(\underbrace{1, 1, \dots, 1, 1}_{2^n \ \text{ones}})$. The $(1, 0, 0, 1)$ is a similar. Reflect to get $(1, 0, 1, 1, 0, 1)$, which then ends up repeating into $(1, 0, \underbrace{1, 1, \dots, 1, 1}_{2^n - 4 \ \text{ones}}, 0, 1)$. $\blacksquare$ Here's the first two reflections shown: [asy][asy] import geometry; size(5cm); int W=4; int H=3; for (int i=-W; i<=W; ++i) { draw( (i,-H)--(i,H), rgb("333333")+dotted ); } for (int i=-H; i<=H; ++i) { draw( (-W,i)--(W,i), rgb("333333")+dotted ); } pen pinkL = pink+white+white; pen purpleL = purple+white+white; filldraw(shift(2,1)*unitsquare, purpleL); filldraw(shift(2,-1)*unitsquare, purpleL); filldraw(shift(-1,-1)*unitsquare, purpleL); filldraw(shift(-1,1)*unitsquare, purpleL); filldraw(shift(2,0)*unitsquare, pinkL); filldraw(shift(2,-2)*unitsquare, pinkL); filldraw(shift(-1,-2)*unitsquare, pinkL); filldraw(shift(-1,0)*unitsquare, pinkL); filldraw(shift(0,1)*unitsquare, purpleL+blue+purpleL); filldraw(shift(0,-1)*unitsquare, purpleL+blue+purpleL); filldraw(shift(-3,-1)*unitsquare, purpleL+blue+purpleL); filldraw(shift(-3,1)*unitsquare, purpleL+blue+purpleL); filldraw(shift(0,0)*unitsquare, pinkL+blue+pinkL); filldraw(shift(0,-2)*unitsquare, pinkL+blue+pinkL); filldraw(shift(-3,-2)*unitsquare, pinkL+blue+pinkL); filldraw(shift(-3,0)*unitsquare, pinkL+blue+pinkL); draw(line((0,0),(1,0)), red+dashed); draw(line((0,0),(0,1)), red+dashed); [/asy][/asy]

I overlooked the $(1, 0, 0, 1)$ case and thus died on this problem for an embarassing long period of time. In hindsight, the fact that there's no good way of proving impossibility should've been a sign of the opposite. It's not hard to convince yourself that it is impossible tough; if you trick yourself into believing that a construction is only possible if the reflected figure is continguous at some point in time, it's probably joever. By considering adjacent $L$ trominos, it's not hard to prove that all reflections must be at multiples of $45^\circ$ wrt the orientation of the base tromino. I want to say this is 40 MOHS but I mostly just skill issued majorly lmao.